Q1. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Given,
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 = 5/40 = 0.125 kg / m
We know,
Velocity of the transverse wave , v =
 |
Q2. A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top, given that the speed of sound in air is 340 m s–1? (g = 9.8 m s–2)
Answer:
Given,
Height of the bridge, s = 300 m
The initial velocity of the stone, u = 0
Acceleration, a = g = 9.8 m/s2
Speed of sound in air = 340 m/s
Let t be the time taken by the stone to hit the water’s surface
We know,
s = ut + ½ gt2
300 = 0 + ½ x 9.8 x t2
Therefore, t = 7.82 s
The time taken by the sound to reach the bridge, t’= 300/340 = 0.88 s
Therefore, from the moment the stone is released from the bridge, the sound of it splashing the water is heard after = t +t’ = 7.82 s + 0.88 s = 8.7s.
Q3. Steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1.
Answer:
Given,
Length of the steel wire, l = 12 m
Mass of the steel wire, m = 2.0 kg
Velocity of the transverse wave, v = 343 m/s
Mass per unit length, μ = M /l = 2.10/12 = 0.175 kg / m
We know,
Velocity of the transverse wave , v =
= 3432 x 0.175 = 2.06 x 104 N
Therefore, the tension in the wire is 3432 x 0.175 = 2.06 x 104 N.
Q5.We know that the function y = f (x, t) represents a wave travelling in one direction, where x and t must appear in the combination x + vt or x– vt or, i.e. y = f (x ± vt). Is the converse true?
Can the following functions for y possibly represent a travelling wave:
(i) (x – vt) 2
(ii) log [ ( x + vt)/ x0 ] (iii) 1 / (x + vt )
Answer:
No, the converse is not true because it is necessary for a wave function representing a travelling wave to have a finite value for all values of x and t.
As none of the above functions satisfies the given condition, none of the options represents a travelling wave.
Q6. A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound and (b) the transmitted sound? The speed of sound in air is 340 m s –1, and in water, 1486 m s–1.
Answer:
Given,
Frequency of the ultrasonic sound, ν = 1000 kHz = 106 Hz
Speed of sound in air, vA = 340 m/s
We know,
(a) The wavelength ( λR ) of the reflected sound is:
λR = vA /v
= 340/106 = 3.4 x 10-4 m
(b) Speed of sound in water, vW = 1486 m/s
Therefore, the wavelength (λT )of the transmitted sound is:
λT = 1486 / 106
= 1.49 x 10-3 m.
Q7. An ultrasonic scanner operating at 4.2 MHz is used to locate tumours in tissues. If the speed of sound is 2 km/s in a certain tissue, calculate the wavelength of sound in this tissue.
Answer:
Given,
Speed of sound in the tissue, vT = 2 km/s = 2 × 103 m/s
Operating frequency of the scanner, ν = 4.2 MHz = 4.2 × 106 Hz
Therefore, the wavelength of sound:
λ = vT / v
= (2 × 103)/ (4.2 × 106)
= 4.76 x 10-4 m.
Q8. A transverse harmonic wave on a wire is expressed as:
y( x, t ) = 3 sin ( 36t +0.018x +Ï€/4 ).
(i) Is it a stationary wave or a travelling one?
(ii) If it is a travelling wave, give the speed and direction of its propagation.
(iii) Find its frequency and amplitude.
(iv) Give the initial phase at the origin.
(v) Calculate the smallest distance between two adjacent crests in the wave.
[X and y are in cm and t in seconds. Assume the left to the right direction as the positive direction of x]
Answer:
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 /2Ï€ = 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.
Q9. For the wave in the above question (Q8), plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm.
(i) Give the shapes of these plots.
(ii) With respect to which aspects (amplitude, frequency or phase) does the oscillatory motion in a travelling wave differ from one point to another?
Answer:
(i) 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
=> t = π/18 secs.
Plotting the displacement (y) vs (t) graphs using different values of t listed below:
Q10. A travelling harmonic wave is given as: y (x, t) = 2.0 cos 2Ï€ (10t – 0.0080x + 0.35).
What is the phase difference between the oscillatory motion of two points separated by a distance of:
(i) 8 m
(ii) 1 m
(iii) λ /2
(iv) 6λ/4
[ X and y are in cm, and t is in secs ].
Answer:
Given,
The equation for a travelling harmonic wave:
y (x, t) = 2.0 cos 2Ï€ (10t – 0.0080x + 0.35)
= 2.0 cos (20Ï€t – 0.016Ï€x + 0.70 Ï€)
Where,
Propagation constant, k = 0.0160 π
Amplitude, a = 2 cm
Angular frequency, ω= 20 π rad/s
We know,
Phase difference Φ = kx = 2π/ λ
(i) For x = 8 m = 800 cm
Φ = 0.016 Ï€ × 800 = 12.8 Ï€ rad.
(ii) For x = 1 m = 100 cm
Φ = 0.016 Ï€ × 100 = 1.6 Ï€ rad.
(iii) For x = λ /2
Φ = (2π/ λ) x ( λ/2) = π rad.
(iv) For x = 6λ /4
Φ = (2π/ λ) x ( 6λ/4) = 3 π rad.
Q11. The transverse displacement of a wire (clamped at both its ends) is described as:
y (x, t) =
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)
=
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