First slide

Simple hormonic motion

Question

A frictionless inclined surface having inclination $30^{0}$ has a 2.5 kg mass held by a spring which is fixed at the upper end. If the mass is taken 2.5 cm up along the surface, tension in the spring reduces to zero. On releasing the mass now, its angular frequency of oscillations will be:

Moderate

A

9
B

14
C

2
D

5

Solution

In equilibrium, $Mg \sin 30^{o} = kx$
$\Rightarrow k = \frac{Mg son 30^{o}}{x} k = \frac{2.5 \times 9.8 \times \frac{1}{2}}{2.5 \times 10^{- 2}} = 4.9 \times 10^{2} N / m ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{4.9 \times 10^{2}}{2.5}} = 14$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App