Question

A smooth inclined plane having an angle of inclination ${30}^{\circ }$ with horizontal has a mass $2.5kg$ held by a spring which is fixed at the upper end of the inclined plane, if the mass is taken $2.5cm$ up along the surface of the inclined plane from the equilibrium position, the tension in the spring reduces to zero. if the mass is then released, the angular frequency of oscillation in radians per second is

• A. $0.707$
• B. $7.07$
• C. $1.414$
• D. $14.4$

Answer 1

The correct option is D $14.4$

The downward acceleration along inclined plane

$T=2\pi \sqrt{\frac{m}{k}}T=2\pi \sqrt{\frac{l}{g^1}}$

$|$ m = mass of particle

K = spring constant of particle

g' = change acceleration

l = elongation by mass $|$

Now as displacement up along plane by 2.5 cm makes tension zero.Hence spring extended by 2.5 cm by mass.

So, angular frequency '$\omega$'

$\omega =\frac{2\pi }{T}\Rightarrow \omega =\sqrt{\frac{g^{\prime }}{l}}\Rightarrow \omega =\sqrt{\frac{g\sin 30°}{2.5}}=\sqrt{\frac{1000\times \frac{1}{2}}{2.5}}\Rightarrow \omega =10\sqrt{2}=14.4$