Alfred Martin

2022-01-04

A block oscillating on a spring has an amplitude of 20 cm. What will the amplitude be if the total energy is doubled? And explain.

Mary Goodson

Beginner2022-01-05Added 37 answers

In the simple harmonic motion, the energy is conserved and the total mechanical energy $E$ has an exprwssion in terms of the force constant $k$ and amplitude $A$ as we have in the equation:

$E=K+U$

$\frac{1}{2}k{A}^{2}=\frac{1}{2}m{v}^{2}+\frac{1}{2}k{x}^{2}$

The total mechanical energy:

$E=\frac{1}{2}k{A}^{2}$

$A=\sqrt{\frac{2E}{k}}$

As we have in the equation 1, the amplitude is directly propotional to square root of$E$

$A=\sqrt{E}$

As$k$ is constant, we can find the amplitude for two instants when $E$ is doubled by

$\frac{{A}_{1}}{{A}_{2}}=\frac{\sqrt{E}}{\sqrt{2E}}$

$\frac{{A}_{1}}{{A}_{2}}=\frac{1}{\sqrt{2}}$

$A}_{2}=\sqrt{2}{A}_{1$

${A}_{2}=\sqrt{2}\left(20cm\right)$

${A}_{2}=28.28cm$

The total mechanical energy:

As we have in the equation 1, the amplitude is directly propotional to square root of

As

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