Question

A block of mass $m=2\text{kg}$ is placed on an inclined smooth surface of inclination $\theta ={60}^{\circ }$. One end of a massless spring of relaxed length $50\text{cm}$ and spring constant $k$ is fixed on top of a plane. A block is attached with another end of spring extends by $2.5\text{cm}$. The mass is displaced slightly and released, the time period of resulting oscillation will be,

• A. $1.17\text{sec}$
• B. $0.17\text{sec}$
• C. $0.34\text{sec}$
• D. $1.54\text{sec}$

Answer 1

The correct option is C $0.34\text{sec}$


At mean position,

$mg\sin \theta =kx$

$\therefore \frac{m}{k}=\frac{x}{g\sin \theta }$

$\therefore$ Time period,

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

$\Rightarrow T=2\pi \sqrt{\frac{x}{g\sin \theta }}$

$\therefore T=2\pi \sqrt{\frac{2.5\times {10}^{-2}}{10\times \sin {60}^{\circ }}}=0.34\text{sec}$

Hence, option $\left(\text{C}\right)$ is correct.

Why this question ? Tip: This question is useful to understand simple harmonic oscillations on inclined plane from Newton's $2^{nd}$ law. By equating forces and restoring forces acts on a body try to find out ratio of mass and force constant in terms of quantities given in problem and calculate time period.