Question

One end of a massless spring of relaxed length 50 cm and spring constant k is fixed on top of a frictionless inclined plane of inclination $\theta ={30}^{\circ }$ as shown in figure. When a mass m = 1.5 kg is attached at the other end, the spring extends by 2.5 cm. The mass is displaced slightly and released. The time period (in seconds) of the resulting oscillation will be

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The force which increases the length of the spring by x = 2.5 cm is F = mg sin $\theta$ Therefore, the spring constant is
$k=\frac{F}{x}=\frac{mgsin\theta }{x}$
Now time period $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{m}{\frac{mgsin\theta }{x}}}=2\pi \sqrt{\frac{x}{gsin\theta }}$
Putting x = 2.5 cm = 2.5 $times$${10}^{-2}m$ , $g=9.8m{s}^{-2}$ and $\theta ={30}^{\circ }$ , we get $T=\frac{\pi }{7}$ second, which is choice (a)