Question

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

• A. $7\text{rad/s}$
• B. $14\text{rad/s}$
• C. $0.7\text{rad/s}$
• D. $1.4\text{rad/s}$

Answer 1

The correct option is B $14\text{rad/s}$


From given data, we can conclude that
Extension in the spring at equilibrium, $x=2.5\text{cm}$
Hence,
$2.5g\sin {30}^{\circ }=kx$

$\Rightarrow k=\frac{2.5g\sin {30}^{\circ }}{2.5\times {10}^{-2}}=\frac{2.5\times 9.8\times \left(1/2\right)}{2.5\times {10}^{-2}}$

$\Rightarrow k=4.9\times {10}^2\text{N/m}$

Angular frequency, $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{4.9\times {10}^2}{2.5}}=\frac{70}{5}=14\text{rad/s}$

Hence, the correct answer is option (b).