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A 1000 m long rod of density 10.0 xx 10^...

A `1000 m` long rod of density `10.0 xx 10^(4) kg//m^(3)` and having young's modulus `Y=10^(11) Pa`, is clamped at one end. It is hammered at the other free end as shown in the figure. The logitudinal pulse goes to right end, gets reflected and again returns to the left end. How much time (in sec) the pulse take to go back to initial point ?
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The correct Answer is:
2
Velocity of longitudinal `u=sqrt((Y)/(rho))=sqrt((10^(11))/(10xx10^(4)))=10^(3)ms^(-1)`
Required time `(2l)/(v)=(2xx1000)/(10^(3))=2s`.
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