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

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

'''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?'''

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

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

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

'''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?'''

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

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

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

$$E_e \propto \cos^2 \omega t$$

$$E_k \propto \sin^2 \omega t$$



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.

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

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

Substitute these into the equation for the total energy:

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

We know that:

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

Therefore:

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

By substitution:

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