Question

The fundamental frequency of a longitudinal vibrating of a rod clamped at its centre is 1500 Hz. If the mass of the rod is 96 g the increase in its total length produced by a stretching force of 10 kg weight will be:

Solution

Correct option is

0.011 cm

 

The wavelength of waves in the rod = 2l (l = length of rod).

                    

Since the vibrations are longitudinal,

           

Since V = Al (where A is the area of c.s.)

If x is the increase in length produced by 10 kg weight, it follows:

        

from eq. (2) and (3),

        

Solving, we get x = 0.011 cm.

 

SIMILAR QUESTIONS

Q1

What is the ratio of the speed of sound in neon and water vapours at the same temperature? It is nearest to:

Q2

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Q3

A man beats a drum at a certain distance from a mountain. He slowly increases the rate of being and finds that the echo is not heard distinctly when the drum beating is at the rate of 40 per minute. He moves by 80 m towards the mountain and finds that the echo is again not heard distinctly when the rate of beating of the drum is 1 per second. What is the original distance of the man from the mountain?

Q4

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Q5

A slab of mass m is released from a height h0 from the top of a spring S’ of force constant  K. The maximum compression x of the spring is given by the equation:

                                                                           

Q6

A bob of mass m is oscillating as a simple pendulum with maximum amplitude θm. What is the maximum tension in the string?

Q7

A certain amount of acoustic energy is emitted in a uniform hemispherical pattern from a point source. At a distance of 5 m form the source the intensity is 5 W/m2. At a distance of 15 m the intensity will be:

Q8

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Q9

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Q10

If there are three sources of sound of equal intensity with frequencies 300, 301 and 302, the number of beats heard per second will be: