Advertisements
Advertisements
प्रश्न
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 m s–1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s–1? The speed of sound in still air can be taken as 340 m s–1.
Advertisements
उत्तर १
The travelling harmonic wave is y(x, t) = `7.5 sin (0.0050x+ 12t + pi/4)`
At x = 1 cm and t = 1 sec
`y (1, 1) = 7.5 sin (0.005 xx 1 + 12 xx 1 pi/4) = 7.5 sin (12.005 + pi/4)` ....(i)
Now `theta = (12.005 + pi/4) "radian"`
`= 180/pi (12.005 + pi/4) "degree" = (12.005 xx 180)/(22/7) + 45 = 732.55^@`
`:. From (i), y(1,2) = 7.5 sin (732.55^@) = 7.5 sin (720 + 12.55^@)`
`= 7.5 sin 12.55^@ = 7.5 xx 0.2173 = 1.63 cm`
Velocity of oscillation `v = (dy)/(dt) (1,1) = d/dt[7.5 sin(0.005x + 12 + pi/4)]`
`= 7.5 xx 12 cos [0.005x + 12t + pi/4]`
At x = 1 cm, t = 1 sec
`v = 7.5 xx 12 cos (0.005 + 12 + pi/4) = 90 cos (732.35^@)`
= 90 cos (720 + 12.55)
`v = 90 cos (12.55^@) = 90 xx 0.9765 = 87.89 "cm/s"`
Comparing the given equation with the standard from `y(x,t)= t sin[pi/4 (vt + x) + phi_0]`
We get `r = 7.5 cm, (2piv)/lambda = 12 or 2 piv = 12 `
`v = 6/pi`
`(2pi)/lambda` = 0.005
`:. lambda = 2pi/0.005 = (2xx3.14)/0.005 = 1256 cm = 12.56 m`
Velocity of wave propagation, `v = vlambda = 6/pi xx 12.56 = 24 "m/s"`
We find that velocity at x = 1 cm, t = 1 sec is not equal tovelocity of waves propagation.
b) Now all point which are at a distance of `+- lambda +-2lambda, +- 3lambda` from x = 1 cm will have same tranverse displacement and velocity.As lambda = 12.56 m, therefore all points at distances `+- 12.6 m, +- 25.2 m, +- 37.8 m` from x = 1 cm will have same dispalcement and velocity as at x=1 point t =2 s, 5s and 11s
उत्तर २
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, ve = 340 + 10 = 350 m/s
The wavelength (λ) of the sound heard by the observer is given by the relation:
`lambda = V_e/V` = 350/400 = 0.875 m`
For the running observer:
Velocity of the observer, vo = 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_0)/v)v`
`= ((340+ 10)/340) xx 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.
APPEARS IN
संबंधित प्रश्न
A bat emits an ultrasonic sound of frequency 1000 kHz in the air. If the sound meets a water surface, what is the wavelength of the transmitted sound? The speed of sound in air is 340 m s–1 and in water 1486 m s–1.
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h–1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s–1.
Show that for a wave travelling on a string
\[\frac{y_{max}}{\nu_{max}} = \frac{\nu_{max}}{\alpha_{max}},\]
where the symbols have usual meanings. Can we use componendo and dividendo taught in algebra to write
\[\frac{y_{max} + \nu_{max}}{\nu_{max} - \nu_{max}} = \frac{\nu_{max} + \alpha_{max}}{\nu_{max} - \alpha_{max}}?\]
A sine wave is travelling in a medium. A particular particle has zero displacement at a certain instant. The particle closest to it having zero displacement is at a distance
Two wave pulses travel in opposite directions on a string and approach each other. The shape of one pulse is inverted with respect to the other.
Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is 120°. The resultant amplitude will be
The equation of a wave travelling on a string stretched along the X-axis is given by
\[y = A e {}^- \left( \frac{x}{a} + \frac{t}{T} \right)^2 .\]
(a) Write the dimensions of A, a and T. (b) Find the wave speed. (c) In which direction is the wave travelling? (d) Where is the maximum of the pulse located at t = T? At t = 2 T?
A pulse travelling on a string is represented by the function \[y = \frac{a^2}{\left( x - \nu t \right)^2 + a^2},\] where a = 5 mm and ν = 20 cm-1. Sketch the shape of the string at t = 0, 1 s and 2 s. Take x = 0 in the middle of the string.
A wave pulse is travelling on a string with a speed \[\nu\] towards the positive X-axis. The shape of the string at t = 0 is given by g(x) = Asin(x/a), where A and a are constants. (a) What are the dimensions of A and a ? (b) Write the equation of the wave for a general time t, if the wave speed is \[\nu\].
A wave propagates on a string in the positive x-direction at a velocity \[\nu\] \[t = t_0\] is given by \[g\left( x, t_0 \right) = A \sin \left( x/a \right)\]. Write the wave equation for a general time t.
A string of length 20 cm and linear mass density 0⋅40 g cm−1 is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t = 0 near an ends as shown in the figure, which travels towards the other end. (a) When will the string have the shape shown in the figure again? (b) Sketch the shape of the string at a time half of that found in part (a).
A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is 1⋅0 and the displacement becomes zero 200 times per second. The linear mass density of the string is 0⋅10 kg m−1 and it is kept under a tension of 90 N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10 ms.
The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by
\[y = \left( 0 \cdot 4 cm \right) \sin\left[ \left( 0 \cdot 314 {cm}^{- 1} \right) x \right] \cos \left[ \left( 600\pi s^{- 1} \right) t \right]\]
(a) What is the frequency of vibration? (b) What are the positions of the nodes? (c) What is the length of the string? (d) What is the wavelength and the speed of two travelling waves that can interfere to give this vibration?
A 40 cm wire having a mass of 3⋅2 g is stretched between two fixed supports 40⋅05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is 1⋅0 mm2, find its Young modulus.
A string 1 m long is fixed at one end. The other end is moved up and down with a frequency of 20 Hz. Due to this, a stationary wave with four complete loops gets produced on the string. Find the speed of the progressive wave which produces the stationary wave.
For the travelling harmonic wave
y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)
Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of 4 m.
A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36t + 0.018x + π/4) where x and y are in cm and t is in s. The positive direction of x is from left to right.
- The wave is travelling from right to left.
- The speed of the wave is 20 m/s.
- Frequency of the wave is 5.7 Hz.
- The least distance between two successive crests in the wave is 2.5 cm.
At what temperatures (in °C) will the speed of sound in air be 3 times its value at O°C?
The amplitude of wave disturbance propagating in the positive x-direction given is by `1/(1 + x)^2` at time t = 0 and `1/(1 + (x - 2)^2)` at t = 1 s, where x and y are in 2 metres. The shape of wave does not change during the propagation. The velocity of the wave will be ______ m/s.
