# how to solve A body is executing SHM with amplitude A and time period T. The ratio of kinetic energy when displacement from equilibrium position is half the amplitude and 1/4th amplitude is

Answers (1)

Amplitude = A

Total Energy of SHM = E =  $\frac{1}{2}kA^{2}$

Let the spring constant be k

Potetial energy when displacement from equilibrium position is x $=\frac{1}{2}kx^{2}$

Potetial energy when displacement from equilibrium position is half the amplitude = U1

$\\U_{1}=\frac{1}{2}k\left ( \frac{A}{2} \right )^{2}\\ U_{1}=\frac{kA^{2}}{8}$

Kinetic energy when displacement from equilibrium position is half the amplitude = K1 = E - U1

$\\K_{1}=\frac{1}{2}kA^{2}-\frac{1}{8}kA^{2}\\ K_{1}=\frac{3}{8}kA^{2}$

Potetial energy when displacement from equilibrium position is one-fourth the amplitude = U2

$\\U_{2}=\frac{1}{2}k\times \left ( \frac{A}{4} \right )^{2}\\ U_{2}=\frac{1}{32}kA^{2}$

Kinetic energy when displacement from equilibrium position is one-fourth the amplitude = K2 = E - U2

$\\K_{2}=\frac{1}{2}kA^{2}-\frac{1}{32}kA^{2}\\ K_{2}=\frac{15}{32}kA^{2}$

$\\\frac{K_{1}}{K_{2}}=\frac{\frac{3}{8}kA^{2}}{\frac{15}{32}kA^{2}}\\ \frac{K_{1}}{K_{2}}=\frac{4}{5}$

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