Sound Wave:Displacement Equation
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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 cm 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.
(All quantities are in SI units.)
- 40 m
- 20 m
- 60 m
- 80 m
A car moves with a speed of 54 km h−1 towards a cliff. The horn of the car emits sound of frequency 400 Hz at a speed of 335 m s−1. (a) Find the wavelength of the sound emitted by the horn in front of the car. (b) Find the wavelength of the wave reflected from the cliff. (c) What frequency does a person sitting in the car hear for the reflected sound wave ? (d) How many beats does he hear in 10 seconds between the sound coming directly from the horn and that coming after the reflection ?
- 500 Hz
- 1000 Hz
- 1200 Hz
- 1500 Hz
A train approaching a platform at a speed of 54 km/h sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platfrom ? The speed of sound in air = 332 m/s.
- Ratio of displacement amplitude of the particles to the wavelength of wave is 1.1×10−5.
- Ratio of displacement amplitude of the particles to the wavelength of wave is 1.59×10−5.
- Ratio of the velocity amplitude of the particles to the wave speed is 2×10−4.
- Ratio of the velocity amplitude of the particles to the wave speed is 10−4.
- Ratio of displacement amplitude to wavelength of wave is 3.86×10−4.
- The velocity amplitude of the particle is 36×10−2 m/s.
- Ratio of the velocity amplitude of the particle to the wave speed is 2.43×10−3.
- The velocity amplitude of the particle is 3.6×10−2 m/s.
One end of a long string of linear mass density 8.0×10−3 kg m−1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
(Consider velocity of sound in air as 1100 ft/s)
- Lowest frequency heard by the listener is 515 Hz
- Lowest frequency heard by the listener is 525 Hz
- Highest frequency heard by the listener is 545 Hz
- Highest frequency heard by the listener is 555 Hz
- Zero
- π2
- π
- All of these
- 10
- 0.5
- 1
- 0.11
- 312 W
- 512 W
- 412 W
- 312 W
- f2=f0(v2(v−v0)(2v−v0))
- f2=2f0(v2(v−v0)(2v−v0))
- f2=f0(v(v+v0)(v−v0)(2v−v0))
- f2=2f0(v(v+v0)(v−v0)(2v−v0))
- 1950Hz
- 1832Hz
- 1650Hz
- 2068Hz
- two
- One
- Four
- Eight
- 1616
- 21.37
- 1200
- 92.7
- 7
- 6
- 5
- 3.5
- 1.33
- 1.40
- 1.50
- 1.6
- λ12
- λ8
- λ24
- λ16
- ω
- 2ω
- ω2
- ω4
(Also assume that the wave propagates along the positive x-direction).
Equation of displacement of a point 2.5m from the child's end can be expressed as
- y=−(0.1m)sin(10rad/s)t
- y=(0.1m)sin(4rad/s)t
- y=−(0.1m)cos(18.8rad/s)t
- y=0.1mcos(12.5rad/s)t
- 0.63 W
- 0.83 W
- 0.47 W
- 0.89 W
- λ4
- λ3
- λ2
- λ
- 10sec-1
- 2sec-1
- 1sec-1
- 0.01sec-1