Question

If for a particle of mass 10 g executing SHM along a straight line, the time period is 2 s and amplitude is 10 cm then what will be its kinetic energy when it is at 5 cm from its equilibrium position?

• A. $ 0.375\,{{\pi }^{2}}\,\text{erg} $
• B. $ 3.75\,{{\pi }^{2}}\,\text{erg} $
• C. $ 37.5\,{{\pi }^{2}}\,\text{erg} $
• D. $ 375\,{{\pi }^{2}}\,\text{erg} $

Answer 1

Answer: (D)

Kinetic energy of a particle executing SHM is $ K=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}}) $ $ =\frac{1}{2}\times 10{{\left( \frac{2\pi }{2} \right)}^{2}}[{{(10)}^{2}}-{{(5)}^{2}}] $ $ =375\,{{\pi }^{2}}\,\text{erg} $