Question

One end of a massless spring of natural length $50cm$ and spring constant $K$ is fixed on top of a frictionless inclined plane $\theta ={30}^{\circ }$ as shown. When a mass of $1.5kg$ is attached to the pre and it extends the spring by $2.5cm$ in equilibrium. The frequency of oscillations will be:

• A. $\frac{5\sqrt{2}}{2\pi }Hz$
  
• B. $\frac{5\sqrt{2}}{\pi }Hz$
• C. $\frac{7\sqrt{2}}{\pi }Hz$
  
• D. $\frac{7\sqrt{2}}{2\pi }Hz$
  

Answer 1

The correct option is B $\frac{5\sqrt{2}}{\pi }Hz$

GIven, mass $m=1.5kg$
Distance of spring extended, $y=2.5cm$
Angle of inclined plane $\theta ={30}^{\circ }$
This ring is extends $2.5cm$ when mass is attached.
$k_y=mg\sin \theta$
$k_y=\frac{mg\sin \theta }{y}$
$K=\frac{1.5\times 10\times \sin {30}^{\circ }}{2\cdot 5\times {10}^{-2}}$
$K=\frac{1.5\times 10\times \frac{1}{2}}{2.5\times {10}^{-2}}$
$K=300N/m$
FRequency of the oscillation,
$f=\frac{1}{2\pi }\sqrt{\frac{K}{m}}$
$f=\frac{1}{2\pi }\sqrt{\frac{300}{1.5}}$
$f=\frac{1}{2\pi }\sqrt{200}$
$f=\frac{10\sqrt{2}}{2\pi }$
$f=\frac{10\sqrt{2}}{2\pi }Hz$
Final answer : (a)