Advertisements
Advertisements
प्रश्न
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Advertisements
उत्तर १
Ultrasonic beep frequency emitted by the bat, ν = 40 kHz
Velocity of the bat, vb = 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)) xx 40`
`= 40/0.97` kHz`
This frequency is reflected by the stationary wall (`v_s = 0`) 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) xx 40/0.97`
`= (1.03 xx 40)/ 0.97 = 42.47 kHz`
उत्तर २
Here, the frequency of sound emitted by the bat, υ = 40 kHz. Velocity of bat, υs = 0.03 υ, where υ is velocity of sound. Apparent frequency of sound striking the wall
v' = v/(v - v_s) xx v = v/(v - 0.03v) xx 40 kHz
= `40/0.97 kHz`
This frequency is reflected by the wall and is received by the bat moving towards the wall
So `v_s = 0`
`v_L = 0.03 v`
`v" = (v+v_L)/v xx v'= (v + 0.03v)/v (40/0.97)`
`= 1.03/0.97 xx 40 kHz = 42.47 kHz`
`:. y = (1,1) = 7.5 sin (732.81^@)`
`= 7.5 sin (90 xx 8 + 12.81^@) = 7.5 sin 12.81^@`
= 7.5 xx 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 [7.5 sin(0.0050x + 12t + pi/4)]`
`= 7.5 xx 12 cos (0.0050x + 12t + pi/4)`
At x = 1 cm and t = 1 s.
`v = y(1,1)= 90 cos(0.0050x + 12t + pi/4)`
At x = 1 cm and t = 1 s
`v = y(1,1) = 90 cos (12.005 + pi/4)`
= 90 cos(732.81^@) = 90 cos(90xx8+ 12.81)
`= 90 cos(12.81^@)`
`= 90 xx 0.975 = 87.75 "cm/s"`
Now the equation of a propagating wave is given by
`y(x,t) = a sin(kx + wt + phi)`
Where
k = (2pi)/lambda
`:. lambda = (2pi)/k`
And `omega = 2piv`
`:. v = omega/(2pi)`
`Speed, v = vlambda = omega/k`
संबंधित प्रश्न
Explain why (or how) In a sound wave, a displacement node is a pressure antinode and vice versa,
An open organ pipe of length L vibrates in its fundamental mode. The pressure variation is maximum
Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2 : 1 the radii are in the ratio 3 : 1 and the densities are in the ratio 1 : 2. Find the ratio of their fundamental frequencies.
A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m s−1 and the amplitude is 0⋅5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the X-axis along the string with the origin at one end and t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.
Two wires of same material are vibrating under the same tension. If the first overtone of first wire is equal to the second overtone of second wire and radius of first wire is twice the radius of the second then the ratio of length of first wire to second wire is
Water waves produced by a motor boat sailing in water are ______.
The displacement of a string is given by y (x, t) = 0.06 sin (2πx/3) cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10−2 kg.
- It represents a progressive wave of frequency 60 Hz.
- It represents a stationary wave of frequency 60 Hz.
- It is the result of superposition of two waves of wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m/s in opposite direction.
- Amplitude of this wave is constant.
The wave pattern on a stretched string is shown in figure. Interpret what kind of wave this is and find its wavelength.
The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is 360 ms–1 and their frequencies are 256 Hz.
- Calculate the time at which the second curve is plotted.
- Mark nodes and antinodes on the curve.
- Calculate the distance between A′ and C′.
A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water (Figure). The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, maximum intensity of sound is heard. If the room temperature is 20°C, calculate
- speed of sound in air at room temperature
- speed of sound in air at 0°C
- if the water in the tube is replaced with mercury, will there be any difference in your observations?
Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1:2:3:4.
Two travelling waves produce a standing wave represented by the equation. y = 1.0 mm cos (1.57 cm-1) x sin (78.5 s-1)t. The node closest to the origin in the region x > 0 will be at x = ______ cm.
Two closed end pipes when sounded together produce 5 beat per second. If their length are in the ratio 100 : 101, then fundamental notes produced by them are ______.
A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ______ cm.
(Take the speed of sound in air as 340 ms-1.)
A tuning fork of frequency 480 Hz is used in an experiment for measuring the speed of sound (ν) in the air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l1 = 30 cm and l2 = 70 cm. Then, ν is equal to ______.
A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is ______.
