Question

A mass is suspended from a vertical spring which is executing S.H.M of frequency $5$ Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is [acceleration due to gravity g = $10m/s^2$]

• A. $2\pi m/s$
• B. $\pi m/s$
• C. $\frac{1}{2\pi }m/s$
• D. $\frac{1}{\pi }m/s$

Answer 1

The correct option is D $\frac{1}{\pi }m/s$

We know that block has maximum speed at equilibrium position.

By applying conservation of mechanical energy at equilibrium position and highest point of oscillation:-

Energy at equilibrium=energy at highest point

$\frac{1}{2}m{v}^2+\frac{1}{2}k{A}^2=mgA$ (speed at highest point and elongation in spring is $0$.)

And , $\omega =2\pi f=10\pi s^{-1}$

$\omega =\sqrt{\frac{k}{m}}⟹k=m{\omega }^2$

At equilibrium:-

$mg=kA⟹A=\frac{mg}{k}=\frac{g}{{\omega }^2}$

Now:-

$\frac{1}{2}m{v}^2+\frac{1}{2}m{\omega }^2{A}^2=mgA$

$⟹v^2=2gA-{\omega }^2{A}^2$

$⟹v^2=\frac{2g^2}{{\omega }^2}-\frac{g^2}{{\omega }^2}=\frac{g^2}{{\omega }^2}$

$⟹v^2=\frac{{10}^2}{(10\pi {)}^2}=\frac{1}{{\pi }^2}$

$⟹v=\frac{1}{\pi }m/s$

Hence, answer is option-(D).