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 isA body is executing SHM with amplitude A and time period T. The

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S Sayak

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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