Worked example 11.2: Block and spring

Question: A block attached to a spring executes simple harmonic motion in a horizontal plane with an amplitude of $0.25 {\rm m}$. At a point $0.15 {\rm m}$ away from the equilibrium position, the velocity of the block is $0.75 {\rm m/s}$. What is the period of oscillation of the block?

Answer: The equation of simple harmonic motion is

$$x = a \cos(\omega t-\phi),$$

where $x$ is the displacement, and $a$ is the amplitude. We are told that $a=0.25 {\rm m}$. The velocity of the block is obtained by taking the time derivative of the above expression:

$$\dot{x} =-a \omega \sin(\omega t -\phi).$$

We are told that at $t=0$ (say), $x = 0.15 {\rm m}$ and $\dot{x} = 0.75 {\rm m/s}$. Hence,

$$0.15 = 0.25 \cos(\phi),$$

$$0.75 = 0.25 \omega \sin(\phi).$$

The first equation gives $\phi = \cos^{-1}(0.15/0.25) = 53.13^\circ$. The second equation yields

$$\omega = \frac{0.75}{0.25\times\sin(53.13^\circ)} = 3.75 {\rm rad./s}.$$

Hence, the period of the motion is

$$T = \frac{2 \pi}{\omega} = 1.676 {\rm s}.$$