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Question

Physics Question on Oscillations

A massless spring of length ll and spring constant kk is placed vertically on a table. A all of mass m is just kept on top of the spring. The maximum velocity of the ball is

A

gmkg \sqrt{\frac{m}{k}}

B

g2mkg \sqrt{\frac{2m}{k}}

C

2gmk2g \sqrt{\frac{m}{k}}

D

g2mk\frac{g}{2} \sqrt{\frac{m}{k}}

Answer

gmkg \sqrt{\frac{m}{k}}

Explanation

Solution

If x=x= displacement of free end of spring, then
mg=kxm g=k x
or X=mgkX=\frac{m g}{k}
Time period of oscillation, T=2πmkT=2 \pi \sqrt{\frac{m}{k}}
ω=\Rightarrow \omega= Angular frequency
=2πT=km=\frac{2 \pi}{T}=\sqrt{\frac{k}{m}}
Maximum velocity of oscillating mass
=Vmax=Aω=V_{\max }=A \omega
=mgk×km=gmk=\frac{m g}{k} \times \sqrt{\frac{k}{m}} =g \sqrt{\frac{m}{k}}