Question

A spring block system (mass $m$, spring constant $k$) is placed on a smooth inclined plane of inclination $\theta$. The inclined plane is accelerated horizontally such that the block does not loose contact with the surface. The time period of small oscillation of the block is

• A. $2\pi \sqrt{\frac{m}{k}}$
• B. $2\pi \sqrt{\frac{msin\theta }{k}}$
• C. $2\pi \sqrt{\frac{mg}{ka}}$
• D. $2\pi \sqrt{\frac{m}{ksin\theta }}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{m}{k}}$

Let $x_0$ be the elongation of spring in equilibrium


$\therefore macos\theta +mgsin\theta =k{x}_o⟶\left(1\right)$

Let $x$ be displacement of mass from equilbrium

$F_{net}=mgsin\theta +macos\theta -k(x_o+x)=-kx$
$\therefore a_{net}=-\frac{kx}{m}$
As $a\propto -{\omega }^{2}x$
mass executes SHM
$\Rightarrow \omega =\sqrt{\frac{k}{m}}$
$\Rightarrow T=2\pi \sqrt{\frac{m}{k}}$

OR

$mg$ and Pseudo - force ($ma$) do not contribute to the restoring force. Hence the time period of the block does not change and $=2\pi \sqrt{\frac{m}{k}}$