📚 FYUG (NEP) previous year question papers solution

ASSAM UNIVERSITY, SILCHAR

FYUG 3rd semester Physics DSC 201 Previous Year Question Papers Solutions

UNIT 2

2019

(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )

1.Find the temperature at which the velocity of sound in air becomes 1.5 times its value at 0°C. (Mark:- 2)

Formula:

v₂ v₁ = √ T₂ T₁

Given:
v₂ = 1.5v₁
T₁ = 273 K

1.5 = √ T₂ 273

Squaring both sides,

2.25 = T₂ 273

T₂ = 2.25 × 273

T₂ = 614.25 K

t = 614.25 − 273

t = 341.25°C

Answer: The required temperature is 341.25°C.

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2.What do you understand by Phase Velocity and Group Velocity? (Mark:- 2)

Formula:

v_p = ω k

v_g = dω dk

Phase Velocity:

Phase velocity is the velocity with which a particular phase of a wave, such as a crest or trough, travels through a medium. It is represented by v_p.

Using the relation:

v_p = ω k

where ω is the angular frequency and k is the wave number.

Group Velocity:

Group velocity is the velocity with which a group of waves or a wave packet moves through a medium. It is represented by v_g.

v_g = dω dk

Group velocity represents the velocity of transmission of energy and information through the medium.

Conclusion:

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3. Show that the frequency of the fundamental note of an open organ pipe is twice that from a closed pipe of the same length. (Mark:- 2)

Formula:

For an Open Organ Pipe:

L = λ₁ 2

f₁ = v λ₁

For a Closed Organ Pipe:

L = λ₂ 4

f₂ = v λ₂

Proof:

Let the length of both pipes be L.

For the open organ pipe,

L = λ₁ 2

λ₁ = 2L

f₁ = v 2L

For the closed organ pipe,

L = λ₂ 4

λ₂ = 4L

f₂ = v 4L

Therefore,

f₁ f₂ = v/2L v/4L

f₁ f₂ = 4L 2L

f₁/f₂ = 2

Therefore,

f₁ = 2f₂

Answer: The frequency of the fundamental note of an open organ pipe is twice the frequency of a closed organ pipe of the same length.

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4.Starting from the relation
v = √ E ρ
for velocity of sound in a gas, show that
v = √ γP ρ
where P is the pressure and γ is the ratio of specific heat at constant pressure to specific heat at constant volume. (Mark:- 4)

for velocity of sound in a gas, show that

v = √ γP ρ

4. Starting from the relation

v = √ E ρ

for velocity of sound in a gas, show that

v = √ γP ρ

where P is the pressure and γ is the ratio of specific heat at constant pressure to specific heat at constant volume.

Given the relation

v = √ E ρ

Laplace assumed that propagation of sound in a gas is an adiabatic process.

PV^γ = Constant

Consider a gas having pressure P and volume V.

If its pressure increases by a small amount ΔP and volume decreases by a small amount ΔV, then

PV^γ = (P + ΔP)(V − ΔV)^γ

Writing,

PV^γ = (P + ΔP) V^γ ( 1 - ΔV V )^γ

Dividing both sides by V^γ,

P = (P + ΔP) ( 1 - ΔV V )^γ

Using Binomial Theorem,

(1 - x)^γ = 1 - γx

Therefore,

( 1 - ΔV V )^γ = 1 - γ ΔV V

Substituting,

P = (P + ΔP) [ 1 - γ ΔV V ]

Multiplying,

P = P - γP ΔV V + ΔP - γΔP ΔV V

Since

γΔP ΔV V

is very small, it can be neglected.

0 = - γP ΔV V + ΔP

ΔP = γP ΔV V

Now,

E = Stress Strain

E = ΔP ΔV/V

Substituting the value of ΔP,

E = γP

Substituting in

v = √ E ρ

we get,

v = √ γP ρ

Hence, the velocity of sound in a gas is

v = √ γP ρ

Hence Proved.

5.Obtain the expression for phase velocity and group velocity in terms of angular frequency and propagation number. (Mark:- 4)

Let a simple harmonic progressive wave be represented by

y = a sin (ωt − kx)

where,

a = amplitude of vibration
ω = angular frequency
k = propagation number (wave number)
x = distance travelled by the wave
t = time

To obtain the phase velocity, let

φ = ωt − kx

where φ is the phase of the wave.

For a particular phase,

φ = Constant

Differentiating with respect to time,

dφ/dt = 0

ω − k dx dt = 0

Therefore,

ω = k dx dt

dx dt = ω k

But

v_p = dx dt

Hence phase velocity is

v_p = ω k

Now we derive the expression for group velocity.

Consider the superposition of two waves of equal amplitude having slightly different frequencies and wave numbers.

y₁ = A cos(k₁x − ω₁t)

y₂ = A cos(k₂x − ω₂t)

Resultant displacement,

y = y₁ + y₂

y = A cos(k₁x − ω₁t) + A cos(k₂x − ω₂t)

Using the relation

cos C + cos D = 2 cos C + D 2 cos C − D 2

we get,

y = 2A cos (k₁+k₂)x − (ω₁+ω₂)t 2 cos (k₁−k₂)x − (ω₁−ω₂)t 2

Let,

ω = ω₁ + ω₂ 2

k = k₁ + k₂ 2

Δω = ω₁ − ω₂

Δk = k₁ − k₂

Then,

y = 2A cos Δkx − Δωt 2 cos(kx − ωt)

The amplitude of the resultant wave is

A(x,t) = 2A cos Δkx − Δωt 2

The group travels with the velocity

v_g = Δω Δk

When Δω and Δk become very small,

v_g = dω dk

Hence the expression for phase velocity is

v_p = ω k

and the expression for group velocity is

v_g = dω dk

Hence proved.

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6. Calculate the energy of the 5th vibration of a stretched string plucked at h, the initial displacement of the plucked point being k.

Let,

L = Length of the string
h = Distance of the plucked point from one end
k = Initial displacement of the plucked point
s = Mode number
T = Tension in the string
μ = Mass per unit length of the string

When the string is plucked, the initial shape of the string is triangular and the displacement is given by

y(x,0) = kx h for 0 ≤ x ≤ h

y(x,0) = k(L-x) L-h for h ≤ x ≤ L

The Fourier expansion of a plucked string gives the amplitude of the s-th mode as

A_s = 2k s²π² [ sin(sπh/L) h/L ]

Simplifying,

A_s = 2kL s²π²h sin ( sπh L )

The energy of the s-th mode of vibration of a stretched string is

E_s = 1 2 T ( sπ L )² A_s²

Substituting the value of A_s,

E_s = 1 2 T ( sπ L )² ( 2kL s²π²h sin ( sπh L ) )²

Squaring the bracket,

E_s = 1 2 T ( s²π² L² ) ( 4k²L² s⁴π⁴h² ) sin² ( sπh L )

Cancelling common terms,

E_s = 2Tk² s²π²h² sin² ( sπh L )

For the 5th vibration,

s = 5

Therefore,

E₅ = 2Tk² 25π²h² sin² ( 5πh L )

Hence, the energy of the 5th vibration of the stretched string is

E₅ = 2Tk² 25π²h² sin² ( 5πh L )

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7. Describe Melde's experiment and explain how laws of vibration of strings can be verified with this experiment.

Melde's Experiment:

Melde's experiment is used to study the transverse vibrations of a stretched string and to verify the laws of vibrating strings.

In this experiment, one end of a light string is attached to a prong of an electrically maintained tuning fork. The string passes over a frictionless pulley and a weight is suspended from the free end. The suspended weight provides the tension in the string.

When the tuning fork vibrates, periodic transverse waves are produced in the string. Under suitable conditions of tension and length, stationary waves are formed in the string. The string then vibrates in a number of loops separated by nodes.

If the string vibrates in p loops, then

L = p λ 2

Therefore,

λ = 2L p

The frequency of vibration of the string is

n = v λ

Substituting the value of λ,

n = v 2L/p

n = pv 2L

The velocity of transverse waves in a stretched string is

v = √ T μ

where

T = tension in the string
μ = mass per unit length of the string

Substituting for v,

n = p 2L √ T μ

For the fundamental mode, p = 1.

n = 1 2L √ T μ

This is the fundamental equation of a vibrating string.

Verification of Laws of Vibrating Strings

From the above equation,

n = 1 2L √ T μ

we can verify the three laws of vibrating strings.

1. Law of Length

Keeping tension T and mass per unit length μ constant,

n ∝ 1 L

Thus the frequency is inversely proportional to the vibrating length of the string.

By changing the vibrating length and observing resonance, it is found that when the length increases, the frequency decreases. Hence the law of length is verified.

2. Law of Tension

Keeping length L and mass per unit length μ constant,

n ∝ √T

Thus the frequency is directly proportional to the square root of the tension.

By varying the suspended load and hence the tension, it is found that the frequency varies as √T. Therefore the law of tension is verified.

3. Law of Mass

Keeping length L and tension T constant,

n ∝ 1 √μ

Thus the frequency is inversely proportional to the square root of the mass per unit length.

Using strings of different materials and thicknesses, it is found that heavier strings produce lower frequencies. Hence the law of mass is verified.

Combining all three laws,

n = 1 2L √ T μ

Thus Melde's experiment successfully verifies the laws of vibrating strings.

Conclusion: The frequency of a stretched string is inversely proportional to its length, directly proportional to the square root of tension, and inversely proportional to the square root of mass per unit length. These laws are verified experimentally by Melde's experiment.

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2022

(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )

8. Distinguish between Stationary and Progressive Waves.

Stationary Waves	Progressive Waves
Formed by superposition of two identical waves travelling in opposite directions.	Produced by a source and travel through the medium in one direction.
Energy is not transferred from one point to another.	Energy is continuously transferred through the medium.
Nodes and antinodes are formed.	No nodes and antinodes are formed.
Different particles vibrate with different amplitudes.	All particles have the same amplitude in a uniform medium.
Wave profile does not move forward.	Wave profile moves continuously.
Phase difference between particles varies from point to point.	Phase changes continuously along the direction of propagation.

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9. What is phase velocity? Why is it also called wave velocity?

Phase velocity is the velocity with which a particular phase of a wave, such as a crest, trough or any fixed point on the wave, travels through the medium.

The phase velocity is given by

v_p = ω k

where,
ω = angular frequency
k = propagation number (wave number)

It is called wave velocity because it represents the speed with which the wave pattern or phase propagates through the medium.

Thus phase velocity gives the velocity of propagation of the wave itself.

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10. What is group velocity? Under what condition group velocity is equal to the phase velocity?

Group velocity is the velocity with which a group of waves or wave packet travels through a medium.

It represents the velocity of transmission of energy and information.

The group velocity is given by

v_g = dω dk

where,
ω = angular frequency
k = propagation number

The phase velocity is

v_p = ω k

Group velocity becomes equal to phase velocity in a non-dispersive medium, where the phase velocity is independent of wavelength or frequency.

v_g = v_p

Hence, in a non-dispersive medium the group velocity and phase velocity are equal.

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11. Derive the expression for Newton's formula for velocity of sound. Explain Laplace's correction.

According to Newton, the propagation of sound through a gas is an isothermal process.

The velocity of sound in a medium is given by

v = √ E ρ

where
E = Bulk modulus of elasticity
ρ = Density of the gas

For an isothermal process,

PV = Constant

Differentiating,

P dV + V dP = 0

V dP = - P dV

dP dV/V = -P

Therefore, bulk modulus

E = P

Substituting in the velocity equation,

v = √ P ρ

This is Newton's formula for velocity of sound in a gas.

Laplace's Correction

Newton assumed that the compressions and rarefactions produced during sound propagation are isothermal. This assumption was incorrect because these changes occur very rapidly and there is no time for heat exchange.

Laplace suggested that the propagation of sound is an adiabatic process.

PV^γ = Constant

For an adiabatic process,

E = γP

Substituting in the velocity equation,

v = √ γP ρ

This is Laplace's corrected formula for the velocity of sound in gases.

Hence Newton's formula was corrected by replacing P with γP.

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12. If the velocity of sound in hydrogen be 1300 m/s at a certain temperature, what will be the velocity at the same temperature in a diatomic gas of molecular weight 32?

For gases at the same temperature,

v ∝ √ γ M

Therefore,

v₁ v₂ = √ γ₁M₂ γ₂M₁

For hydrogen,

v₁ = 1300 m/s
M₁ = 2
γ₁ = 1.4

For the diatomic gas,

M₂ = 32
γ₂ = 1.4

1300 v₂ = √ 1.4 × 32 1.4 × 2

1300 v₂ = √16

1300 v₂ = 4

v₂ = 1300 4

v₂ = 325 m/s

Answer: Velocity in the diatomic gas = 325 m/s.

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13(a). Find the expression for group velocity and hence find the relation between group velocity and phase velocity.

The group velocity is defined as the velocity with which a group of waves or wave packet travels.

v_g = dω dk

The phase velocity is

v_p = ω k

Therefore,

ω = kv_p

Differentiating with respect to k,

dω dk = v_p + k dv_p dk

But

v_g = dω dk

Hence,

v_g = v_p + k dv_p dk

This is the relation between group velocity and phase velocity.

For a non-dispersive medium,

dv_p dk = 0

Therefore,

v_g = v_p

13. (b) What is meant by dispersive medium?

A dispersive medium is a medium in which the velocity of a wave depends on its frequency or wavelength.

In such a medium, waves of different frequencies travel with different velocities.

Hence the phase velocity varies with wavelength and

v_g ≠ v_p

Examples: Water waves, light waves in glass, and electromagnetic waves in optical fibres.

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2023

(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )

14. What modification did Laplace make in Newton's assumption to calculate the velocity of sound correctly?

Newton assumed that the compressions and rarefactions produced during the propagation of sound in a gas are isothermal processes.

PV = Constant

On this basis, Newton obtained

v = √ P ρ

However, the calculated value was lower than the experimental value.

Laplace pointed out that the compressions and rarefactions occur very rapidly and there is no time for heat exchange with the surroundings.

Therefore, the process is adiabatic and not isothermal.

PV^γ = Constant

Hence the bulk modulus becomes

E = γP

Substituting in the velocity equation,

v = √ γP ρ

This correction is known as Laplace's correction.

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15. Explain briefly the concept of stationary wave.

A stationary wave is formed by the superposition of two waves of the same frequency, wavelength and amplitude travelling in opposite directions in the same medium.

Let the two waves be

y₁ = a sin(ωt − kx)

y₂ = a sin(ωt + kx)

The resultant displacement is

y = y₁ + y₂

y = 2a sinωt coskx

This equation represents a stationary wave.

In a stationary wave, certain points remain permanently at rest and are called nodes, while points vibrating with maximum amplitude are called antinodes.

The distance between two consecutive nodes or antinodes is

λ 2

and the distance between a node and the adjacent antinode is

λ 4

No energy is transferred from one point to another in a stationary wave.

Examples of stationary waves are vibrations of stretched strings and air columns in organ pipes.

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16. Show that
y = a sin ( 2π λ (l − x) )
satisfies the wave equation.

Given,

y = a sin ( 2π λ (l − x) )

Let

k = 2π λ

Then,

y = a sin k(l − x)

Differentiating partially with respect to x,

∂y ∂x = -ak cos k(l − x)

Again differentiating with respect to x,

∂²y ∂x² = -ak² sin k(l − x)

Since

y = a sin k(l − x)

therefore

∂²y ∂x² = -k²y

The one-dimensional wave equation is

∂²y ∂t² = v² ∂²y ∂x²

Substituting

∂²y ∂x² = -k²y

we obtain

∂²y ∂t² = -v²k²y

which satisfies the standard wave equation.

Hence

y = a sin ( 2π λ (l − x) )

is a valid solution of the wave equation.

Hence proved.

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17. Derive the expression for the velocity of transverse vibrations of a stretched string.

Consider a stretched string under tension T. Let μ be the mass per unit length of the string.

Suppose a transverse wave travels along the string with velocity v. Consider a small element PQ of the string of length δx.

Let the tensions at P and Q be T and make angles θ₁ and θ₂ with the horizontal.

Since the displacement is small,

T_P = T_Q = T

The horizontal components cancel each other.

The resultant vertical force on the element is

F = T sinθ₂ - T sinθ₁

F = T(sinθ₂ - sinθ₁)

For small angles,

sinθ ≈ tanθ

Hence,

F = T ( ∂y ∂x )_x+δx - T ( ∂y ∂x )_x

F = T ∂²y ∂x² δx

Mass of the element

m = μδx

Acceleration of the element

a = ∂²y ∂t²

By Newton's second law,

F = ma

T ∂²y ∂x² δx = μδx ∂²y ∂t²

Cancelling δx,

T ∂²y ∂x² = μ ∂²y ∂t²

∂²y ∂t² = T μ ∂²y ∂x²

Comparing with the standard wave equation,

∂²y ∂t² = v² ∂²y ∂x²

Therefore,

v² = T μ

v = √ T μ

Hence the velocity of transverse vibrations of a stretched string is

v = √ T μ

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18. Explain the differences between progressive and standing waves with examples.

Progressive Waves	Standing Waves
Travel continuously through the medium.	Do not travel through the medium.
Energy is transmitted from one point to another.	No net transfer of energy occurs.
No nodes and antinodes are formed.	Nodes and antinodes are formed.
All particles have nearly the same amplitude.	Amplitude varies from point to point.
Phase changes continuously with position.	Particles between two nodes vibrate in the same phase.
Produced by a single travelling wave.	Produced by superposition of two identical waves travelling in opposite directions.

Examples of Progressive Waves:

1. Sound waves travelling in air.
2. Ripples moving on the surface of water.
3. Light waves travelling through space.

Examples of Standing Waves:

1. Vibrations of a stretched string fixed at both ends.
2. Air columns in organ pipes.
3. Vibrations in Melde's experiment.

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19. Consider a stretched string fixed at both ends. Derive the expression for standing wave pattern on the string. Show that both even and odd harmonics are present.

Consider two progressive waves of equal amplitude and frequency travelling in opposite directions along a stretched string.

y₁ = a sin(ωt - kx)

y₂ = a sin(ωt + kx)

The resultant displacement is

y = y₁ + y₂

y = a sin(ωt-kx)+a sin(ωt+kx)

Using the relation

sinC + sinD = 2 sin C+D 2 cos C-D 2

we get

y = 2a sinωt coskx

This is the equation of a stationary wave.

Since the string is fixed at both ends,

y = 0

x = 0

and

x = L

where L is the length of the string.

At x = L,

2a sinωt coskL = 0

Since

sinωt ≠ 0

therefore

coskL = 0

kL = (2n+1)π 2

where

n = 0,1,2,3,...

Since

k = 2π λ

therefore

2πL λ = (2n+1)π 2

λ = 4L (2n+1)

The frequency is

f = v λ

f = (2n+1)v 4L

For different values of n,

n = 0

f₁ = v 4L

n = 1

f₃ = 3v 4L

n = 2

f₅ = 5v 4L

Thus only odd harmonics are present in a string fixed at one end and free at the other.

For a stretched string fixed at both ends,

L = n λ 2

Therefore

λ = 2L n

Frequency,

f = v λ

f = nv 2L

For

n = 1

f₁ = v 2L

n = 2

f₂ = 2v 2L = v L

n = 3

f₃ = 3v 2L

n = 4

f₄ = 4v 2L

Hence the frequencies are

f₁ , 2f₁ , 3f₁ , 4f₁ , ...

Therefore both even and odd harmonics are present in a stretched string fixed at both ends.

Result: The stationary wave equation is

y = 2a sinωt coskx

and a stretched string fixed at both ends contains both even and odd harmonics.

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2024

(FYUG PHY-DSC201 )

20. What are transverse waves and longitudinal waves?

Transverse Waves:

A transverse wave is a wave in which the particles of the medium vibrate perpendicular to the direction of propagation of the wave.

Examples:

1. Waves on a stretched string.
2. Electromagnetic waves.

Longitudinal Waves:

A longitudinal wave is a wave in which the particles of the medium vibrate parallel to the direction of propagation of the wave.

Examples:

1. Sound waves in air.
2. Compression waves in a spring.

21. Define phase velocity and group velocity.

Phase Velocity:

Phase velocity is the velocity with which a particular phase of a wave, such as a crest or trough, travels through a medium.

v_p = ω k

where ω is the angular frequency and k is the wave number.

Group Velocity:

Group velocity is the velocity with which a group of waves or wave packet travels through a medium.

v_g = dω dk

Group velocity represents the velocity of transmission of energy and information.

22. What do you mean by standing waves? Give one example.

Standing waves or stationary waves are produced by the superposition of two waves of the same frequency, wavelength and amplitude travelling in opposite directions in the same medium.

The resultant wave does not travel from one place to another and hence no energy is transferred.

In a standing wave, nodes and antinodes are formed.

Example:

Stationary waves produced in a stretched string fixed at both ends.

y = 2a sinωt coskx

23. Show that the velocity of a plane progressive wave in a string is given by

v = √ T ρ

where,

T = tension in the string
ρ = linear density (mass per unit length) of the string.

Consider a transverse wave travelling along a stretched string under tension T.

Take a small element PQ of the string of length δx.

Let the tensions at P and Q make angles θ₁ and θ₂ respectively with the horizontal.

Since the displacement is small, the tension throughout the string may be taken as constant.

T_P = T_Q = T

The horizontal components of tension cancel each other.

The resultant vertical force acting on the element is

F = T sinθ₂ - T sinθ₁

F = T (sinθ₂ - sinθ₁)

For small angles,

sinθ ≈ tanθ

Therefore,

F = T (tanθ₂ - tanθ₁)

But,

tanθ = ∂y ∂x

Hence,

F = T [ ( ∂y ∂x )_x+δx - ( ∂y ∂x )_x ]

F = T ∂²y ∂x² δx

Mass of the element PQ is

m = ρδx

Acceleration of the element is

a = ∂²y ∂t²

Applying Newton's second law,

F = ma

T ∂²y ∂x² δx = ρδx ∂²y ∂t²

Cancelling δx from both sides,

T ∂²y ∂x² = ρ ∂²y ∂t²

∂²y ∂t² = T ρ ∂²y ∂x²

Comparing with the standard wave equation

∂²y ∂t² = v² ∂²y ∂x²

we obtain

v² = T ρ

Taking square root on both sides,

v = √ T ρ

Hence, the velocity of a plane progressive wave in a stretched string is

v = √ T ρ

Hence proved.

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24. Derive Newton's formula for velocity of sound. What was its limitation? How did Laplace make the correction?

According to Newton, sound travels through a gas by successive compressions and rarefactions.

The velocity of sound in a medium is given by

v = √ E ρ

where,

E = Modulus of elasticity of the medium
ρ = Density of the medium

Newton assumed that compressions and rarefactions take place isothermally.

PV = Constant

Differentiating,

PdV + VdP = 0

VdP = -PdV

dP dV = - P V

Bulk modulus is

E = - V dP dV

Substituting the value of dP/dV,

E = P

Therefore,

v = √ P ρ

This is Newton's formula for the velocity of sound in a gas.

Limitation of Newton's Formula

Newton assumed that the propagation of sound is an isothermal process.

Using this formula, the velocity of sound in air at 0°C comes out to be about 280 m/s, whereas the experimental value is about 332 m/s.

Thus Newton's formula gave a value smaller than the observed value.

The error arose because compressions and rarefactions occur very rapidly and there is no time for heat exchange with the surroundings.

Laplace's Correction

Laplace pointed out that the propagation of sound is an adiabatic process rather than an isothermal process.

PV^γ = Constant

Differentiating,

PγV^γ-1dV + V^γdP = 0

γP dV V + dP = 0

dP = -γP dV V

Bulk modulus is

E = - V dP dV

E = γP

Substituting in the velocity equation,

v = √ γP ρ

This is Laplace's corrected formula for the velocity of sound in gases.

Hence, Newton's formula was corrected by replacing P with γP.

v = √ γP ρ

Hence proved.

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25. Explain analytically the formation of standing wave in a string fixed at both ends. Hence show how odd and even harmonics are present in such case.

Consider two progressive waves of equal amplitude and frequency travelling in opposite directions along a stretched string.

y₁ = a sin(ωt − kx)

y₂ = a sin(ωt + kx)

The resultant displacement is

y = y₁ + y₂

y = a sin(ωt − kx) + a sin(ωt + kx)

Using the trigonometric identity

sinC + sinD = 2 sin C + D 2 cos C − D 2

we get

y = 2a sinωt coskx

This equation represents a standing wave.

The amplitude of vibration is

A = 2a coskx

For nodes,

A = 0

coskx = 0

kx = (2n+1)π 2

For antinodes,

A = ±2a

coskx = ±1

kx = nπ

Thus nodes and antinodes are formed alternately on the string.

Since the string is fixed at both ends, the ends must be nodes.

x = 0

x = L

At x = L,

sin(kL) = 0

kL = nπ

Since

k = 2π λ

therefore

2πL λ = nπ

λ = 2L n

The frequency of vibration is

f = v λ

f = nv 2L

For n = 1,

f₁ = v 2L

For n = 2,

f₂ = 2v 2L = v L

For n = 3,

f₃ = 3v 2L

For n = 4,

f₄ = 4v 2L

Hence the frequencies are

f₁ , 2f₁ , 3f₁ , 4f₁ , ...

Therefore both even harmonics (2f₁, 4f₁, ...) and odd harmonics (3f₁, 5f₁, ...) are present.

Hence the stationary wave formed on a string fixed at both ends contains both odd and even harmonics.

Result:

y = 2a sinωt coskx

is the equation of the standing wave and both odd and even harmonics are present.

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26. Derive the relation between phase velocity and group velocity.

The phase velocity is defined as the velocity with which a particular phase of a wave travels through the medium.

v_p = ω k

where

ω = angular frequency
k = propagation constant (wave number)

Therefore,

ω = kv_p

Differentiating both sides with respect to k,

dω dk = d dk (kv_p)

Using the product rule,

dω dk = v_p + k dv_p dk

But group velocity is defined as

v_g = dω dk

Substituting,

v_g = v_p + k dv_p dk

This is the relation between group velocity and phase velocity.

Now,

k = 2π λ

Therefore,

dk = - 2π λ² dλ

dv_p dk = - λ² 2π dv_p dλ

Substituting in the above equation,

v_g = v_p - k λ² 2π dv_p dλ

Since

k = 2π λ

Hence,

v_g = v_p - λ dv_p dλ

Therefore, the relation between group velocity and phase velocity is

v_g = v_p + k dv_p dk

v_g = v_p - λ dv_p dλ

For a non-dispersive medium,

dv_p dλ = 0

Therefore,

v_g = v_p

Hence proved.

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CBCS PYQ SAME AS FYUG PHYSICS DSC 201

CBCS PHYHCC-202 Question Paper 2019	Click Here
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FYUG PYQ PHYSICS DSC 201

FYUG PHYSICS DSC 201 Question Paper 2024

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ASSAM UNIVERSITY, SILCHAR

UNIT 2

2019

(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )

1.Find the temperature at which the velocity of sound in air becomes 1.5 times its value at 0°C. (Mark:- 2)

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2.What do you understand by Phase Velocity and Group Velocity? (Mark:- 2)

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3. Show that the frequency of the fundamental note of an open organ pipe is twice that from a closed pipe of the same length. (Mark:- 2)

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4.Starting from the relation v = √ E ρ for velocity of sound in a gas, show that v = √ γP ρ where P is the pressure and γ is the ratio of specific heat at constant pressure to specific heat at constant volume. (Mark:- 4)

4. Starting from the relation

for velocity of sound in a gas, show that

where P is the pressure and γ is the ratio of specific heat at constant pressure to specific heat at constant volume.

5.Obtain the expression for phase velocity and group velocity in terms of angular frequency and propagation number. (Mark:- 4)

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6. Calculate the energy of the 5th vibration of a stretched string plucked at h, the initial displacement of the plucked point being k.

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7. Describe Melde's experiment and explain how laws of vibration of strings can be verified with this experiment.

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2022

(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )

8. Distinguish between Stationary and Progressive Waves.

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9. What is phase velocity? Why is it also called wave velocity?

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10. What is group velocity? Under what condition group velocity is equal to the phase velocity?

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11. Derive the expression for Newton's formula for velocity of sound. Explain Laplace's correction.

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12. If the velocity of sound in hydrogen be 1300 m/s at a certain temperature, what will be the velocity at the same temperature in a diatomic gas of molecular weight 32?

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13(a). Find the expression for group velocity and hence find the relation between group velocity and phase velocity.

13. (b) What is meant by dispersive medium?

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2023

(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )

14. What modification did Laplace make in Newton's assumption to calculate the velocity of sound correctly?

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15. Explain briefly the concept of stationary wave.

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16. Show that y = a sin ( 2π λ (l − x) ) satisfies the wave equation.

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17. Derive the expression for the velocity of transverse vibrations of a stretched string.

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18. Explain the differences between progressive and standing waves with examples.

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19. Consider a stretched string fixed at both ends. Derive the expression for standing wave pattern on the string. Show that both even and odd harmonics are present.

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2024

(FYUG PHY-DSC201 )

20. What are transverse waves and longitudinal waves?

21. Define phase velocity and group velocity.

22. What do you mean by standing waves? Give one example.

23. Show that the velocity of a plane progressive wave in a string is given by

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24. Derive Newton's formula for velocity of sound. What was its limitation? How did Laplace make the correction?

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4.Starting from the relation
v = √ E ρ
for velocity of sound in a gas, show that
v = √ γP ρ
where P is the pressure and γ is the ratio of specific heat at constant pressure to specific heat at constant volume. (Mark:- 4)

16. Show that
y = a sin ( 2π λ (l − x) )
satisfies the wave equation.