A block of mass 1,, kg is hanging vertically | vedclass

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Question

Speed of a transverse wave on a string is given by the formula$:$

$v=\sqrt{\frac{T}{\mu}}$$..........(1)$

where, $T=$ tension in the string

$

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

Speed of a transverse wave on a string is given by the formula$:$

$v=\sqrt{\frac{T}{\mu}}$$..........(1)$

where, $T=$ tension in the string

$\mu=$ mass per unit length (linear density)

In this question$:$

Tension in the string $(T)=m g \ldots \ldots \ldots \ldots . .$ (Newtons Laws of motion) $T=1 	imes 10=10 N$

Mass per unit length $(\mu)=0.001 \mathrm{kg} / \mathrm{m}$

Substituting the respective values in the formula $(1)$

$v=\sqrt{\frac{10}{0.001}}=100 \mathrm{m} / \mathrm{s}$

Hence, the time taken by the generated pulse to reach the other end at a distance of $1 \mathrm{m}$ from the lower end$:$

$t=\frac{s}{v}$

Here, $s=1 m \quad$ and $\quad v=100 m / s e c$

$t=\frac{1}{100}$

$t=0.01$ seconds

A block of mass $1\,\, kg$ is hanging vertically from a string of length $1\,\, m$ and mass /length $= 0.001\,\, Kg/m$. A small pulse is generated at its lower end. The pulse reaches the top end in approximately .... $\sec$

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