# A-level Physics (Advancing Physics)/Energy in Simple Harmonic Motion/Worked Solutions

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1. A 10g mass causes a spring to extend 5 cm. How much energy is stored by the spring?

Gravitational Potential Energy transferred completely to Elastic Potential Energy

Gravitational Potential Energy = mass x gravity x height

GPE ${\displaystyle =0.01m\times 0.05kg\times 9.8Nkg^{-1}=4.9\times 10^{-3}J}$

2. A 500g mass on a spring (k=100) is extended by 0.2m, and begins to oscillate in an otherwise empty universe. What is the maximum velocity which it reaches?

${\displaystyle {\frac {1}{2}}mv_{max}^{2}={\frac {1}{2}}kx_{max}^{2}}$

${\displaystyle v_{max}^{2}={\frac {kx_{max}^{2}}{m}}}$

${\displaystyle v_{max}=x_{max}{\sqrt {\frac {k}{m}}}=0.2\times {\sqrt {\frac {100}{0.5}}}=2.83{\mbox{ ms}}^{-1}}$

3. Another 500g mass on another spring in another otherwise empty universe is extended by 0.5m, and begins to oscillate. If it reaches a maximum velocity of 15ms−1, what is the spring constant of the spring?

${\displaystyle {\frac {1}{2}}mv_{max}^{2}={\frac {1}{2}}kx_{max}^{2}}$

${\displaystyle k={\frac {mv_{max}^{2}}{x_{max}^{2}}}={\frac {0.5\times 15^{2}}{0.5^{2}}}=450{\mbox{ Nm}}^{-1}}$

4. Draw graphs of the kinetic and elastic energies of a mass on a spring (ignoring gravity).

${\displaystyle E_{e}\propto \cos ^{2}\omega t}$

${\displaystyle E_{k}\propto \sin ^{2}\omega t}$

Simple HArmonic Motion

5. Use the trigonometric formulae for x and v to derive an equation for the total energy stored by an oscillating mass on a spring, ignoring gravity and air resistance, which is constant with respect to time.

${\displaystyle x=A\cos {\omega t}}$

${\displaystyle v=-A\omega \sin {\omega t}}$

Substitute these into the equation for the total energy:

${\displaystyle \Sigma E={\frac {1}{2}}(kx^{2}+mv^{2})={\frac {1}{2}}(k(A\cos {\omega t})^{2}+m(-A\omega \sin {\omega t})^{2})={\frac {1}{2}}(kA^{2}\cos ^{2}{\omega t}+mA^{2}\omega ^{2}\sin ^{2}{\omega t})={\frac {A^{2}}{2}}(k\cos ^{2}{\omega t}+m\omega ^{2}\sin ^{2}{\omega t})}$

We know that:

${\displaystyle \omega ={\sqrt {\frac {k}{m}}}}$

Therefore:

${\displaystyle \omega ^{2}={\frac {k}{m}}}$

By substitution:

${\displaystyle \Sigma E={\frac {A^{2}}{2}}(k\cos ^{2}{\omega t}+{\frac {mk}{m}}\sin ^{2}{\omega t})={\frac {A^{2}}{2}}(k\cos ^{2}{\omega t}+k\sin ^{2}{\omega t})={\frac {kA^{2}}{2}}(\cos ^{2}{\omega t}+\sin ^{2}{\omega t})={\frac {kA^{2}}{2}}}$