Advertisements
Advertisements
प्रश्न
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?
उत्तर १
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 `pi/2`
`:. l = lambda/2`
`lambda = 2l = 2x 1 = 2m`
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
उत्तर २
Here, L = 100 cm = 1m, v = 2.53 k Hz = 2.53 x 103 Hz
When the rod is clamped at the middle, then in the fundamental mode of vibration of the rod, a node is formed at the middle and ant mode is formed at each end.
Therefore as is clear from figure
`L= lambda/4 + lambda/4 = lambda/2`
`lambda = 2 L= 2 m`
As `v = vlambda`
`:. v = 2.53 xx 10^3 xx 2`
`= 5.06 xx 10^3 ms^(-1)`
APPEARS IN
संबंधित प्रश्न
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 the particle speed can never be equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by 2π.
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
\[\frac{y_{max} + \nu_{max}}{\nu_{max} - \nu_{max}} = \frac{\nu_{max} + \alpha_{max}}{\nu_{max} - \alpha_{max}}?\]
Velocity of sound in air is 332 m s−1. Its velocity in vacuum will be
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.
Following figure shows two wave pulses at t = 0 travelling on a string in opposite directions with the same wave speed 50 cm s−1. Sketch the shape of the string at t = 4 ms, 6 ms, 8 ms, and 12 ms.
What is the interference of sound waves?
Use the formula `v = sqrt((gamma P)/rho)` to explain why the speed of sound in air is independent of pressure.
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 the reflected sound? The speed of sound in air is 340 m s–1 and in water 1486 m s–1.
Speed of sound wave in air ______.