Question

A pan with a set of weights is attached to a light spring. The period of vertical oscillation is $0.5s$. When some additional weights are put in pan, then the period of oscillations increases by $0.1s$. The extension caused by the additional weight is

• A. $5.5cm$
• B. $3.8cm$
• C. $2.7cm$
• D. $1.3cm$

Answer 1

The correct option is C $2.7cm$

Given, $T=0.5s$,
We know that, $T=2\pi \sqrt{\frac{m}{k}}\Rightarrow 0.5=2\pi \sqrt{\frac{m}{k}}$
$\Rightarrow \frac{1}{4\pi }=\sqrt{\frac{m}{k}}$
On squaring both sides, we get
$\frac{m}{k}=\frac{1}{16{\pi }^2}$ ....(i)
$\Rightarrow T+0.1=2\pi \sqrt{\frac{m+m^{\prime }}{k}}$
$\Rightarrow 0.5+0.1=2\pi \sqrt{\frac{m+m^{\prime }}{k}}$
$\Rightarrow {\left(\frac{0.6}{2\pi }\right)}^2=\frac{m+m^{\prime }}{k}=\frac{m}{k}+\frac{m^{\prime }}{k}$
$\Rightarrow {\left(\frac{0.6}{2\pi }\right)}^2=\frac{1}{16{\pi }^2}+\frac{m^{\prime }}{k}$ [using Equation (i)]
$\Rightarrow \frac{{\left(0.6\right)}^2}{4{\pi }^2}-\frac{1}{16{\pi }^2}=\frac{m^{\prime }}{k}$
$\Rightarrow \frac{m^{\prime }}{k}=\frac{1.44-1}{16{\pi }^2}\Rightarrow \frac{m^{\prime }}{k}=\frac{0.44}{16{\pi }^2}$ .....(ii)
$\because m^{\prime }g=kx$
$\therefore x=\frac{m^{\prime }}{k}g$
$=\frac{0.44}{16{\pi }^2}\times 10$
$=0.0279$
$=2.79\times {10}^{-2}m$
$=2.7cm$