Waves

1. 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,
   The velocity (v) of the transverse wave in the string is given by the relation:
   
   ∴Time taken by the disturbance to reach the other end, t =
2. 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 () taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:
   
   
   
   Time taken by the sound to reach the top of the tower,
   Therefore, the time after which the splash is heard,
   = 7.82 + 0.88 = 8.7 s
3. 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,
   For tension T, velocity of the transverse wave can be obtained using the relation:
   
   ∴T= v²µ
   = (343)²× 0.175 = 20588.575 ≈ 2.06 × 10⁴N
4. (a) Take the relation:
   
   
   Now from the ideal gas equation for n = 1:
   PV = RT
   For constant T, PV = Constant
   Since both M and γare 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:
   
   For one mole of an ideal gas, the gas equation can be written as:
   PV = RT
   P =…(ii)
   Substituting equation (ii) in equation (i), we get:
   
   Where,
   Mass, M = ρVis a constant
   γ and R are also constants
   We conclude from equation (iv) that .
   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) Letbe the speeds of sound in moist air and dry air respectively.
   Letbe the densities of moist air and dry air respectively.
   Take the relation:
   
   Hence, the speed of sound in moist air is:
   
   And the speed of sound in dry air is:
   
   On dividing equations (i) and (ii), we get:
   
   However, the presence of water vapour reduces the density of air, i.e.,
   
   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.
5. Answer: 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
   
   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
   
   Since the function does not converge to a finite value for x = 0 and t = 0, it does not represent a travelling wave.
6. (a) Frequency of the ultrasonic sound, ν = 1000 kHz = 10⁶ Hz
   Speed of sound in air, v_a = 340 m/s
   The wavelength (λ_r) of the reflected sound is given by the relation:
   
   
   (b) Frequency of the ultrasonic sound, ν = 1000 kHz = 10⁶ Hz
   Speed of sound in water, v_w = 1486 m/s
   The wavelength of the transmitted sound is given as:
   = 1.49 × 10^–3 m
7. Speed of sound in the tissue, v = 1.7 km/s = 1.7 × 10³ m/s
   Operating frequency of the scanner, ν = 4.2 MHz = 4.2 × 10⁶ Hz
   The wavelength of sound in the tissue is given as:
   
   
8. Answer:
   (a) Yes; Speed = 20 m/s, Direction = Right to left
   (b) 3 cm; 5.73 Hz
   (c)
   (d) 3.49 m
   Explanation:
   (a) The equation of a progressive wave travelling from right to left is given by the displacement function:
   y (x, t) = a sin (ωt + kx + Φ) … (i)
   The given equation is:
   
   On comparing both the equations, we find that equation (ii) represents a travelling wave, propagating from right to left.
   Now, using equations (i) and (ii), we can write:
   ω = 36 rad/s and k = 0.018 m^–1
   We know that:
   and
   Also,
   v = νλ
   
   Hence, the speed of the given travelling wave is 20 m/s.
   (b) Amplitude of the given wave, a = 3 cm
   Frequency of the given wave:
   
   (c) On comparing equations (i) and (ii), we find that the initial phase angle,
   (d) The distance between two successive crests or troughs is equal to the wavelength of the wave.
   Wavelength is given by the relation:
   
9. All the waves have different phases.
   The given transverse harmonic wave is:
   
   For x= 0, the equation reduces to:
   
   Now, plotting y vs. t graphs using the different values oft, as listed in the given table.
   

   t (s)	0
   y (cm)		3		0		–3		0

   For x= 0, x = 2, and x = 4, the phases of the three waves will get changed. This is because amplitude and frequency are invariant for any change in x. The y-t plots of the three waves are shown in the given figure.
   
    
10. 
    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:
    
    (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
    
    (d) For
    
11. (a) The general equation representing a stationary wave is given by the displacement function:
    y (x, t) = 2a sin kx cos ωt
    This equation is similar to the given equation:
    
    Hence, the given function represents a stationary wave.
    (b) A wave travelling along the positive x-direction is given as:
    
    The wave travelling along the negative x-direction is given as:
    
    The superposition of these two waves yields:
    
    The transverse displacement of the string is given as:
    
    Comparing equations (i) and (ii), we have:
    
    ∴Wavelength, λ = 3 m
    It is given that:
    120π = 2πν
    Frequency, ν = 60 Hz
    Wave speed, v = νλ
    = 60 × 3 = 180 m/s
    (c) The velocity of a transverse wave travelling in a string is given by the relation:
    
    Where,
    Velocity of the transverse wave, v = 180 m/s
    Mass of the string, m = 3.0 × 10^–2kg
    Length of the string, l = 1.5 m
    Mass per unit length of the string,
    
    
    Tension in the string = T
    From equation (i), tension can be obtained as:
    T = v²μ
    = (180)² × 2 × 10^–2
    = 648 N
     
12. 
    Answer:
    (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:
    
    For x = 0.375 m and t = 0
    
13. (a) The given equation represents a stationary wave because the harmonic terms kx and ωtappear 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 ωtare in the combination of kx – ωt.
    (d) The given equation represents a stationary wave because the harmonic terms kx and ωtappear separately in the equation. This equation actually represents the superposition of two stationary waves.
     
14. 
    (a) Mass of the wire, m = 3.5 × 10^–2kg
    Linear mass density,
    Frequency of vibration, ν= 45 Hz
    ∴Length of the wire,
    The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:
    
    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 = νλ=45 × 1.75 = 78.75 m/s
    (b) The tension produced in the string is given by the relation:
    T = v²µ
    = (78.75)² × 4.0 × 10^–2 = 248.06 N
15. Frequency of the turning fork, ν= 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:
    
    Where,
    Length of the pipe,
    
    The speed of sound is given by the relation:
    
16. Length of the steel rod, l = 100 cm = 1 m
    Fundamental frequency of vibration, ν= 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.
    
    The speed of sound in steel is given by the relation:
    v = νλ
    = 2.53 × 10³ × 2
    = 5.06 × 10³ m/s
    = 5.06 km/s
17. Answer: First (Fundamental); No
    Length of the pipe, l = 20 cm = 0.2 m
    Source frequency = n^th normal mode of frequency,ν_n = 430 Hz
    Speed of sound, v = 340 m/s
    In a closed pipe, the n^th normal mode of frequency is given by the relation:
    
    Hence, the first mode of vibration frequency is resonantly excited by the given source.
    In a pipe open at both ends, the n^th mode of vibration frequency is given by the relation:
    
    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.
     
18. 
    Frequency of string A, f_A = 324 Hz
    Frequency of string B = f_B
    Beat’s frequency, n = 6 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:
    
19. (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.
20. (i) (a)Frequency of the whistle, ν= 400 Hz
    Speed of the train, v_T= 10 m/s
    Speed of sound, v = 340 m/s
    The apparent frequency of the whistle as the train approaches the platform is given by the relation:
    
    (b) The apparent frequency of the whistle as the train recedes from the platform is given by the relation:
    
    (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.
21. 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, ν= 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:
    
    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 ().
    This is given by the relation:
    
    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.
22. (a) The given harmonic wave is:
    
    For x= 1 cm and t= 1s,
    
    The velocity of the oscillation at a given point and time is given as:
    
    Now, the equation of a propagating wave is given by:
    
    ∴
    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:
    
    Therefore, all the points at distances nλ,i.e. ±12.56 m, ±25.12 m, … and so on for x= 1 cm, will have the same displacement as thex= 1 cm points at t= 2 s, 5 s, and 11 s.
23. nswer:(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 or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the whistle.
     
24. 
    The equation of a travelling wave propagating along the positivey-direction is given by the displacement equation:
    y(x, t) = a sin (wt – kx) …(i)
    Linear mass density,
    Frequency of the tuning fork, ν = 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:
    
    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
25. Operating frequency of the SONAR system, ν = 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 () received and reflected by the submarine is given by the relation:
    
    The frequency () received by the enemy submarine is given by the relation:
    
     
26. 
    Let v_Sand v_P be the velocities of S andP waves respectively.
    Let Lbe the distance between the epicentre and the seismograph.
    We have:
    L =v_St_S (i)
    L =v_Pt_P (ii)
    Where,
    t_Sand t_P are the respective times taken by theS 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_St_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 × 240 = 480 s
    From equation (ii), we get:
    L = 8 × 240
    = 1920 km
    Hence, the earthquake occurs at a distance of 1920 km from the seismograph.
27. Ultrasonic beep frequency emitted by the bat, ν= 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:
    
    This frequency is reflected by the stationary wall () toward the bat.
    The frequency () of the received sound is given by the relation: