Question

A body of mass 4.5 Kg hangs from a spring and oscillates with a period of 0.9 s. How much will spring shorten when the mass is removed?​

Answer 1

Given:

A body of mass 4.5 kg hangs from a spring and oscillates with a period of 0.9 seconds.

To find:

Shortening of spring after mass is removed?

Calculation:

First of all, let's find out the SPRING CONSTANT of the spring :

$T = 2\pi \sqrt{ \dfrac{m}{k} }$

$\implies \: 0.9 = 2\pi \sqrt{ \dfrac{4.5}{k} }$

$\implies \: \dfrac{0.45}{\pi} = \sqrt{ \dfrac{4.5}{k} }$

$\implies \: \dfrac{4.5}{k} = \dfrac{0.45 \times 0.45}{ {\pi}^{2} }$

$\implies \: \dfrac{1}{k} = 0.00455$

$\implies \: k \approx \: 220 \: N/m$

Now, length of shortening will be :

$F = - kx$

$\implies \: mg = - kx$

$\implies \: 4.5 \times 10 = - 220 \times x$

$\implies \: 45= - 220 \times x$

$\implies \: x = - 4.8 \: m$

• Negative sign shows shortening.

So, spring will be shortened by 4.8 metres.