Question

A smooth inclined plane having angle of inclination ${30}^o$ with horizontal has a mass $2.5$ kg is held by a spring which is fixed at the upper end as shown in figure. If the mass is taken $2.5$cm up along the surface of the inclined plane, the tension in the spring reduces to zero. If the mass is then released, the angular frequency of oscillation in radian per second is?

• A. $0.707$
• B. $7.07$
• C. $1.414$
• D. $14.14$

Answer 1

The correct option is D $14.14$

Tension of string zero means that the string is in unstretched position. So, $2.5cm$ is extension of string when the mass attach to string. From figure, it is clear that the downward acceleration of spring along inclined plane is

$g^{\prime }=g\sin {30}^0=\frac{g}{2}=\frac{10}{2}=5m/s^2$

At the stretched position the restoring force is balanced by weight of mass along inclined plane $k\times x=m{g}^{\prime }=2.5\times 5$

Where, Mass of the body$=2.5kg$

Extension of string$=x=2.5cm=0.025m$

and, $k$=spring constant

$\Rightarrow k\times 0.025=2.5\times 5$

$\Rightarrow k=500$

Angular frequency$\Rightarrow \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{500}{2.5}}$

$\Rightarrow \omega =14.14rad/s$