Question

A 1000 m long rod of density 10.0 x 104 kg/m² and having young's modulus Y = 1011 Pa, is clamped at one
end. It is hammered at the other free end as shown in the figure. The longitudinal pulse goes to right end, gets
reflected and again returns to the left end. How much time (in sec) the pulse will take to go back to initial point?

Answer 1

Answer:

The time taken by the pulse to go back to initial point is 2 s

Explanation:

As we know that the speed of longitudinal wave in a solid medium is given as

$v = \sqrt{\frac{Y}{\rho}}$

here we know that

$Y = 10^{11}$

$\rho = 10 \times 10^4 kg/m^3$

now we have

$v = \sqrt{\frac{10^{11}}{10^5}}$

$v = 1000 m/s$

now time taken by the wave to get back at the same position is given as

$\Delta t = \frac{2L}{v}$

$\Delta t = \frac{2(1000)}{1000}$

$\Delta t = 2 s$

#Learn

Topic : Speed of longitudinal waves

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