Question

If the equation of motion of a particle is given by\(s = A\cos \left( {\omega t + \delta } \right)\), the particle is said to undergo simple harmonic motion.

(a) Find the velocity of the particle at time \(t\).

(b) When is the velocity \(0\)?

Short answer

(a) Thus, the required velocity is \( - \omega A\sin \left( {\omega t + \delta } \right)\).

(b) Thus, the required answer is \(\frac{{n\pi - \delta }}{\omega }\).

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The derivative of the displacement

The derivative of the displacement function of an object is equivalent to the velocity of the object. Therefore, \(v = \frac{{ds}}{{dt}}\).

(a) Step 3: The velocity of the object

Differentiate \(s\) with respect to \(t\) using chain rule as follows:

\(\begin{aligned}\frac{{ds}}{{dt}} &= \frac{d}{{dt}}\left( {A\cos \left( {\omega t + \delta } \right)} \right)\\ &= A\left( { - \sin \left( {\omega t + \delta } \right)} \right)\frac{d}{{dt}}\left( {\omega t + \delta } \right)\\ &= - A\sin \left( {\omega t + \delta } \right)\left( \omega \right)\\ &= - \omega A\sin \left( {\omega t + \delta } \right)\end{aligned}\)

Hence, the velocity of the particle at time \(t\) is \( - \omega A\sin \left( {\omega t + \delta } \right)\).

(b)

The velocity of the particle equal to \(0\) means that \( - \omega A\sin \left( {\omega t + \delta } \right) = 0\). But, \(A \ne 0\) and \(\omega \ne 0\). So, \(\sin \left( {\omega t + \delta } \right) = 0\). Solve it as follows:

\(\begin{aligned}\sin \left( {\omega t + \delta } \right) &= 0\\\omega t + \delta &= n\pi \\\omega t &= n\pi - \delta \\t &= \frac{{n\pi - \delta }}{\omega }\end{aligned}\)

Hence, the velocity is \(0\) at \(t = \frac{{n\pi - \delta }}{\omega }\).