Question

A spring executes SHM with mass of $10 \,kg$ attached to it. The force constant of spring is $10\, N / m$. If at any instant its velocity is $40 \,cm / sec$, the displacement will be (where amplitude is $0.5 \,m$ ) :

• A. 0.09 m
• B. 0.3 m
• C. 0.03 m
• D. 0.9 m

Answer 1

Answer: (B)

0.3 m

Answer 2

Time period of spring is given by $T=2 \pi \sqrt{\left(\frac{m}{k}\right)}$ $\Rightarrow \omega=\frac{2 \pi}{T}=\sqrt{\left(\frac{k}{m}\right)}$ $\therefore \omega=\sqrt{\left(\frac{10}{10}\right)}=1$ Velocity of mass at displacement $x$ $u =\omega \sqrt{a^{2}-y^{2}}$ $\Rightarrow y^{2} =a^{2}-\frac{u^{2}}{\omega^{2}}$ $=(50)^{2}-\frac{(40)^{2}}{1^{2}}$ $=(50)^{2}-(40)^{2}$ $=(50+40)(50-40)=900$ $\therefore y=30\, cm =0.3 \,m$