Question

A mass m is attached to a string of length $l_0$ ,Young’s modulus Y and area of cross section A. It is pushed with a velocity v as shown. Find the time in which the block comes to an instantaneous rest.

• A. $\frac{l_0}{v}+\sqrt{\frac{{\pi }^{2}m{l}_0}{2YA}}$
• B. $\frac{l_0}{v}+\sqrt{\frac{{\pi }^{2}m{l}_0}{4YA}}$
• C. $\frac{l_0}{2v}+\sqrt{\frac{2{\pi }^m{l}_0}{YA}}$
• D. $\frac{l_0}{v}+\sqrt{\frac{2{\pi }^m{l}_0}{YA}}$

Answer 1

The correct option is B $\frac{l_0}{v}+\sqrt{\frac{{\pi }^{2}m{l}_0}{4YA}}$

The string, which is elastic in nature can be replaced by a spring of constant $K=\frac{YA}{l_o}$
Thus

i and f are the extreme positions, which the block travels in a time
$\frac{T}{4}=\frac{\pi }{2}\sqrt{\frac{m}{K}}$
$=\frac{\pi }{2}\sqrt{\frac{m{l}_o}{YA}}$
Thus, the total time taken to reach position of instantaneous rest is
$t=\frac{l_o}{v}+\sqrt{\frac{{\pi }^{2}m{l}_o}{4YA}}$
where $\frac{l_o}{v}:$ time for string to become taut.