Waves Question Answers: NCERT Class 11 Physics

Mass of the string, M = 2.50 kg

Tension in the string, T = 200 N

Length of the string, l = 20.0 m

Mass per unit length, μ = M / l = 2.50 / 20 = 0.125 kg m^-1

The velocity (v) of the transverse wave in the string is given by the relation:

 v  =  underroot (T  / μ)

= Underroot (200 / 0.125) = 40 m/s

∴Time taken by the disturbance to reach the other end, t = l / v  =  20/40 = 0.50s

Height of the tower, s = 300 m

Initial velocity of the stone, u = 0

Acceleration, a = g = 9.8 m/s²

Speed of sound in air = 340 m/s

The time (t₁) taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:

s  =  ut₁  + 1/2 gt²₁

300 =  0 + 1/2 x 9.8 x t12

∴ t₁  =  underroot [(300 x 2)  /  9.8]

= 7.82 s

Time taken by the sound to reach the top of the tower, t₂  =  300/340  = 0.88 s

Therefore, the time after which the splash is heard,  t  =   t₁ +  t₂

= 7.82 + 0.88 = 8.7 s

Length of the steel wire, l = 12 m

Mass of the steel wire, m = 2.10 kg

Velocity of the transverse wave, v = 343 m/s

Mass per unit length, μ = m  /  l  = 2.10  /  12  = 0.175 kg m^-1

For tension T, velocity of the transverse wave can be obtained using the relation:

v  =  underroot ( T / μ)

∴T = v2 µ

= (343)² × 0.175 = 20588.575  ≈  2.06 × 10⁴ N

(a) Take the relation:

v = √ γP/ρ           .... (i)

where density, ρ = Mass / Volume = M / V

Hense eqation one reduces to:

v  =  √ γPV/M                ....(ii)

Now from the ideal gas equation for n = 1:

PV = RT

For constant T, PV = Constant

Since both M and γ are constants, v = Constant

Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.

(b) Take the relation:

v = √ γP/ρ           .... (i)

For one mole of an ideal gas, the gas equation can be written as:

PV = RT 

P = RT / V … (ii)

Substituting equation (ii) in equation (i), we get:

v  =  √ γRT / Vρ =  √γRT / M  ....(iii)

Where, Mass, M = ρV is a constant

 γ and R are also constants

We conclude from equation (iii) that  v  ∝ √T

Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.

(c) Let V_m and V_d be the speeds of sound in moist air and dry air respectively.

Let ρ_m and ρ_d be the densities of moist air and dry air respectively.

Take the relation:

v = √ γP/ρ    

Hence, the speed of sound in moist air is:

v_m = √ γP/ρ_m    .....(i)

And the speed of sound in dry air is:

v_d = √ γP/ρ_d    .....(ii)

On dividing equations (i) and (ii), we get:

However, the presence of water vapour reduces the density of air, i.e.,

ρ_d < ρ_m

therefore, V_m > V_d

Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous medium, the speed of sound increases with humidity.

No;

(a) Does not represent a wave

(b) Represents a wave

(c) Does not represent a wave

The converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should remain finite for all values of x and t.

Explanation:

(a) For x = 0 and t = 0, the function (x - vt)² becomes 0.

Hence, for x = 0 and t = 0, the function represents a point and not a wave.

(b) For x = 0 and t = 0, the function

log [ x  + vt / x₀] = log 0 = ∞

Since the function does not converge to a finite value for x = 0 and t = 0, it represents a travelling wave.

(c) For x = 0 and t = 0, the function

1 /  (x + vt) =  1/0 =  ∞

Since the function does not converge to a finite value for x = 0 and t = 0, it does not represent a travelling wave.

(a) Frequency of the ultrasonic sound, v  = 1000 kHz = 10⁶ Hz

Speed of sound in air, v_a = 340 m/s

The wavelength (v»r) of the reflected sound is given by the relation:

λ_r =  v / V

= 340  / 10⁶ = 3.4 x 10^-4 m

(b) Frequency of the ultrasonic sound, v  = 1000 kHz = 10⁶ Hz

Speed of sound in water, v_w = 1486 m/s

The wavelength of the transmitted sound is given as:

λ_t  =  1486 / 10⁶

= 1.49 × 10^-3 m

Speed of sound in the tissue, v = 1.7 km/s = 1.7 × 10³ m/s

Operating frequency of the scanner, V = 4.2 MHz = 4.2 × 10⁶ Hz

The wavelength of sound in the tissue is given as:

λ =  v / V

= 1.7 × 10³ / 4.2 × 10⁶

= 4.1 x 10^-4 m

(a) Yes; Speed = 20 m/s, Direction = Right to left

(b) 3 cm; 5.73 Hz

(c) π /4

(d) 3.49 m

Explanation:

Given,

y(x, t) =3 sin (36t +0.018x + π/4) . . . . . . . . . . . ( 1 )

( i ) We know, the equation of a progressive wave travelling from right to left is:

y (x, t) = a sin (ωt + kx + Φ) . . . . . . . . . . . . . . . . . . . ( 2 )

Comparing equation ( 1 ) to equation ( 2 ), we see that it represents a wave travelling from right to left and also we get:

a = 3 cm, ω = 36 rad/s , k = 0.018 cm and ϕ = π/4

( ii )Therefore the speed of propagation , v = ω/k = 36/ 0.018 = 20 m/s

( iii ) Amplitude of the wave, a = 3 cm

Frequency of the wave v = ω / 2π = 36 /2π = 5.7 hz

( iv ) Initial phase at the origin = π/4

( v ) the smallest distance between two adjacent crests in the wave, λ = 2π/ k = 2π / 0.018 = 349 cm

Given, y(x, t) =3 sin (36t +0.018x + π/4) . . . . . . . . . . . ( 1 )

For x= 0, the equation becomes :

y( 0, t ) =3 sin ( 36t +0 + π/4 ) . . . . . . . . . . . ( 2 )

Also, ω = 2 π/t = 36 rad/s^-1

therefore t = π/18 secs.

Plotting the displacement (y) vs. (t) graphs using different values of t listed below:

Similarly graphs are obtained for x = 0, x = 2 cm, and x = 4 cm.

The oscillatory motion in the travelling wave is different from each other only in terms of phase. Amplitude and frequency are invariant for any change in x.

The y-t plots of the three waves are shown in the given figure:

Equation for a travelling harmonic wave is given as:

y (x, t) = 2.0 cos 2π (10t - 0.0080x + 0.35)

= 2.0 cos (20πt - 0.016πx + 0.70 π)

Where,

Propagation constant, k = 0.0160 π

Amplitude, a = 2 cm

Angular frequency,  ω = 20 π rad/s

Phase difference is given by the relation:

ø  = kx  =  2 π / λ

(a) For x = 4 m = 400 cm

Φ = 0.016 π × 400 = 6.4 π rad

(b) For 0.5 m = 50 cm

Φ = 0.016 π × 50 = 0.8 π rad

(c) For x  =  λ / 2

Φ =  2 π / λ  x λ / 2 = λ  rad

(d) For x  =  3λ / 4

Φ = 2 π / λ  x 3λ / 4 = 1.5π rad

We know,
The standard equation of a stationary wave is described as:
y (x, t) = 2a sin kx cos ωt

Our given equation y (x, t) =0.06sin(2π/3x)cos(120πt)  is similar to the general equation .
( i ) Thus, the given function describes a stationary wave.

( ii ) We know, a wave travelling in the positive x – direction can be represented as :
y₁ = a sin( ωt – kx )
Also,
Wave travelling in the negative x – direction is represented as :
y₂ = a sin( ωt + kx )

Super- positioning  these two waves gives us :
y = y₁ + y₂
= a sin( ωt – kx ) – a sin( ωt + kx )
= asin(ωt)cos(kx) – asin(kx)cos(ωt) – asin(ωt)cos(kx) – asin(kx)cos(ωt)
= – 2asin(kx)cos(ωt)
= −2asin(2πλx)cos(2πvt) . . . . . . . . . . . . . . ( 1 )

The transverse displacement of the wires is described as :
0.06sin(2π3x)cos(120πt)             . . . . . . . . . . . . . . . . .( 2 )

Comparing equations ( 1 ) and ( 2 ) , we get :
2π/ λ = 2π/ 3
Therefore, wavelength  λ = 3m

Also, 2πv/λ = 120π
Therefore, speed v = 180 m/s

And, Frequency = v/λ = 180/3
= 60 Hz

( iii )Given, Velocity of the transverse wave, v = 180 m / s
The string’s mass, m = 6 × 10^-2 kg
String length, l = 3 m
Mass per unit length of the string, μ = m/l = (6 x 10^-2 )/3
= 2 x 10^-2 kg/m
Let the tension in the wire be T
Therefore, T = v² μ
= 180² x 2 x 10^-2
= 648 N.

(i)

(a) Yes, except at the nodes

(b) Yes, except at the nodes

(c) No

(ii) 0.042 m

Explanation:

(i)

(a) All the points on the string oscillate with the same frequency, except at the nodes which have zero frequency.

(b) All the points in any vibrating loop have the same phase, except at the nodes.

(c) All the points in any vibrating loop have different amplitudes of vibration.

(ii) The given equation is:

 y(x ,t)  =  0.06 sin(2π / 3 x ) cos (120πt)

For x = 0.375 m and t = 0

Amplitude  =  Displacement  = 0.06 sin(2π / 3 x ) cos 0

= 0.06 sin(2π / 3 x 0.0375 )  x 1

= 0.06 sin (0.25π)  =  0.06sin (π /4)

= 0.06 x 1  /  underroot 2

= 0.042 m

(a) The given equation represents a stationary wave because the harmonic terms kx and ωt appear separately in the equation.

(b) The given equation does not contain any harmonic term. Therefore, it does not represent either a travelling wave or a stationary wave.

(c) The given equation represents a travelling wave as the harmonic terms kx and ωt are in the combination of kx - ωt.

(d) The given equation represents a stationary wave because the harmonic terms kx and ωt appear separately in the equation. This equation actually represents the superposition of two stationary waves.

(a) Mass of the wire, m = 3.5 × 10^-2 kg

Linear mass density,  μ =  m / l =  4.0 x 10^-2 kg m^-1

Frequency of vibration, V = 45 Hz

∴Length of the wire,  l  = m / μ =    3.5 × 10^-2  /  4.0 x 10^-2

The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:

λ = 2l / n

where , n = Number of nodes in the wire

For fundamental node, n = 1:

λ = 2l

λ = 2 × 0.875 = 1.75 m

The speed of the transverse wave in the string is given as:

v = V λ = 45 × 1.75 = 78.75 m/s

(b) The tension produced in the string is given by the relation:

T = v² µ

= (78.75)² × 4.0 × 10^-2 = 248.06 N

Frequency of the turning fork, v = 340 Hz

Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.

Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation:

l₁  =  λ / 4

Where, Length of the pipe, l₁ = 25.5 cm = 0.255m

therefore,λ  =  4l₁ = 4 x 0.255  = 1.02 m

The speed of sound is given by the relation:

v  = vλ = 340 × 1.02 = 346.8 m/s

Length of the steel rod, l = 100 cm = 1 m

Fundamental frequency of vibration, V_F = 2.53 kHz = 2.53 × 10³ Hz

When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends, as shown in the given figure.

The distance between two successive nodes is λ / 2.

∴ l = λ / 2

Or, λ = 2 x 1 = 2m

The speed of sound in steel is given by the relation:

v = v = νλ

= 2.53 × 10³ × 2

= 5.06 × 10³ m/s

= 5.06 km/s

First (Fundamental); No

Length of the pipe, l = 20 cm = 0.2 m

Source frequency = n^th normal mode of frequency, V_n = 430 Hz

Speed of sound, v = 340 m/s

In a closed pipe, the nth normal mode of frequency is given by the relation:

V_n  =  (2n -1) v / 4l                    ; n is an integer = 0, 1, 2, 3...

430 = (2n-1) 340 / (4x0.2)

2n -1  =  (430 x 4 x 0.2) / 340 =  1.01

2n = 2.01

n ≈ 1

Hence, the first mode of vibration frequency is resonantly excited by the given source.

In a pipe open at both ends, the nth mode of vibration frequency is given by the relation:

V_n    =  nv  / 2l

n = 2lV_n  / v

 = 2 x 0.2 x 430 / 340  =  0.5

Since the number of the mode of vibration (n) has to be an integer, the given source does not produce a resonant vibration in an open pipe.

Frequency of string A, f_A = 324 Hz

Frequency of string B = f_B

Beat's frequency, n = 6 Hz

Beat's frequency is given as:

n = l f_A +- f_B l

6 =  324 +- f_B

f_B  = 330 Hz or 318 Hz

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:

v ∝ Underroot T

Hence, the beat frequency cannot be 330 Hz

∴ f_B = 318 Hz

(a) A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an antinode is a point where the amplitude of vibration is the maximum and pressure is the minimum.

Therefore, a displacement node is nothing but a pressure antinode, and vice versa.

(b) Bats emit very high-frequency ultrasonic sound waves. These waves get reflected back toward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses.

(c) The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

(d) Solids have shear modulus. They can sustain shearing stress. Since fluids do not have any definite shape, they yield to shearing stress. The propagation of a transverse wave is such that it produces shearing stress in a medium. The propagation of such a wave is possible only in solids, and not in gases.

Both solids and fluids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal waves can propagate through solids and fluids.

(e) A pulse is actually is a combination of waves having different wavelengths. These waves travel in a dispersive medium with different velocities, depending on the nature of the medium. This results in the distortion of the shape of a wave pulse.

(i)

(a) Frequency of the whistle, V = 400 Hz

Speed of the train, v_T= 10 m/s

Speed of sound, v = 340 m/s

The apparent frequency (V') of the whistle as the train approaches the platform is given by the relation:

V' =  (v / v -v_T ) V

= (340 / 340-10 ) x 400 = 412.12 Hz

(b) The apparent frequency (V'')of the whistle as the train recedes from the platform is given by the relation:

V'' =  (v / v + v_T ) V

= (340 / 340 +10 ) x 400 = 388.57 Hz

(ii) The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same, i.e., 340 m/s.

For the stationary observer: 400 Hz; 0.875 m; 350 m/s

For the running observer:  Not exactly identical

For the stationary observer:

Frequency of the sound produced by the whistle, V =  400 Hz

Speed of sound = 340 m/s

Velocity of the wind, v = 10 m/s

As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e., 400 Hz.

The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,

Effective speed of the sound, v_e = 340 + 10 = 350 m/s

The wavelength (λ) of the sound heard by the observer is given by the relation:

λ  =  v_e / V = 350 / 400 = 0.875 m

For the running observer:

Velocity of the observer, v_o = 10 m/s

The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency (v').

This is given by the relation:

v' = (v+v_o / v ) V

= (340+10 / 340) x 400 =  411.76 Hz

Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s

The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875 m.

Hence, the given two situations are not exactly identical.

(a) The given harmonic wave is:

y(x,t) = 7.5sin [0.0050x + 12t + π/4]

For x = 1 cm and t = 1s,

y = (1, 1) = 7.5sin [0.0050 + 12 + π/4]

.= 7.5sin [12.0050 + π/4]

= 7.5 sinθ

Where, θ = 12.0050 + π/4 =  12.0050  + 3.14 / 4 = 12.79 rad

 =  180 /3.14 x 12.79  =  732.81°

∴ y = (1, 1) = 7.5sin [732.81°]

= 7.5 sin (90 x 8 + 12.81°)

= 7.5 sin (12.81°)

= 7.5 x 0.2217

= 1.6629 ≈ 1.663 cm

The velocity of the oscillation at a given point and time is given as:

v = d/dt y(x,t) = d/dt [7.5sin(0.0050x + 12t +π/4)]

= 7.5 x 12cos (0.0050x + 12t +π/4)

At x = 1 cm and t  =  1s:

v =  y(1,1)  =  90 cos (12.005 + π/4)

= 90cos(732.81°) = 90cos(90 x 8 + 12.81°)

= 90cos(12.81°)

= 90 x 0.975 = 87.75 cm/s

Now, the equation of a propagating wave is given by:

y(x,t) = a sin(kx + wt + ø)

Where,

k = 2π / λ

∴  λ =  2π / k

And ω =  2πv

∴ v  =  ω / 2π

Speed  =  v  =  vλ =  ω / k

Where

ω =  12 rad/s

k  =  0.0050 m^-1

∴ v  =  12 /0.0050 = 2400 cm/s

∴ Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:

k  =  2π / λ

∴ λ =  2π / k = 2 x 3.14 / 0.0050

= 1256 cm = 12.56

Therefore, all the points at distances n_λ , (n =±1, ±2....and so on)  i.e. ± 12.56 m, ± 25.12 m, … and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s, and 11 s.

(a) (i)No (ii)No (iii)Yes

(b) No

Explanation: (a) The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that medium.

(b) The short pip produced after every 20 s does not mean that the frequency of the whistle is 1/20 or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the whistle.

The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation:

y (x, t) = a sin (wt - kx) … (i)

Linear mass density, μ =  8.0 x 10^-3 kg m^-1

Frequency of the tuning fork, v = 256 Hz

Amplitude of the wave, a = 5.0 cm = 0.05 m … (ii)

Mass of the pan, m = 90 kg

Tension in the string, T = mg = 90 × 9.8 = 882 N

The velocity of the transverse wave v, is given by the relation:

v  =  underoot T / μ

= underoot 882 / 8.0 x 10^-3 = 332 m/s

Angular Frequency, ω = 2πv

= 2 x 3.14 x 256

= 1608.5 = 1.6 x 10³ rad/s   .....(iii)

Wavelength, λ = v / V = 332 / 256 m

∴ Propagation constant, k = 2π / λ

= 2 x 3.14 / 332/256 = 4.84 m^-1 ......... (iv)

Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:

y (x, t) = 0.05 sin (1.6 × 10³t  - 4.84 x) m

Operating frequency of the SONAR system, v = 40 kHz

Speed of the enemy submarine, v_e = 360 km/h = 100 m/s

Speed of sound in water, v = 1450 m/s

The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparent frequency (v') received and reflected by the submarine is given by the relation:

v' = ( v + v_e / v ) v

= ( 1450+ 100 / 1450 ) x 40 = 42.76 kHz

The frequency (v'') received by the enemy submarine is given by the relation:

v''  = ( v / v + v_s) v'

where, v_s = 100 m/s

∴ v'' = (1450 / 1450-100) x 42.76 = 45.93 kHz

Let v_Sand v_P be the velocities of S and P waves respectively.

Let L be the distance between the epicentre and the seismograph.

We have:

L = v_St_S (i)

L = v_Pt_P (ii)

Where,

t_S and t_P are the respective times taken by the S and P waves to reach the seismograph from the epicentre

It is given that:

v_P = 8 km/s

v_S= 4 km/s

From equations (i) and (ii), we have:

v_S t_S = v_P t_P

4t_S = 8 t_P

t_S = 2 t_P (iii)

It is also given that:

t_S - t_P = 4 min = 240 s

2t_P - t_P = 240

t_P= 240

And t_S = 2 x 240 = 480 s

From equation (ii), we get:

L = 8 x 240

= 1920 km

Hence, the earthquake occurs at a distance of 1920 km from the seismograph.

Ultrasonic beep frequency emitted by the bat, v = 40 kHz

Velocity of the bat, v_b = 0.03 v

Where, v = velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

v' =  [v / v - v_b] v

= [v / v - 0.03v] x 40

= 40/0.97 kHz

This frequency is reflected by the stationary wall (v_s) toward the bat.

The frequency (v'') of the received sound is given by the relation:

v''  =  [ v + v_b / v] v'

= [ v + 0.03v / v] x 40/0.97

= 1.03 x 40  / 0.97 = 42.47 kHz