Question

A clock pendulum oscillates at a frequency of 2.5 Hz. At t= 0, it is released from rest starting at an angle of${12}^{\circ }$to the vertical. Ignoring friction, what will be the position (angle in radians) of the pendulum at (a)t= 0.25 s, (b) t = 1.60 s, and (c) t = 500 s?

Short answer

(a) The position of the pendulum at t = 0.25s will be – 0.15 rad.

(b) The position of the pendulum at t = 1.60 s will be $\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}$.

(c) The position of the pendulum at $t=500$s will be $\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}$.

Step-by-step solution

Understanding the displacement of pendulum

In this problem, firstly determine the relation of angulardisplacement of the pendulum in terms of frequency at any time t and then calculate the position of the pendulum.

Identification of the given information

The frequency of the simple pendulum is f = 2.5 Hz.

The maximum angular displacement of the simple pendulum at t = 0 is, ${\theta }_{max}={12}^{\circ }=\left({12}^{\circ }\right)\left(\frac{\frac{\pi }{180}\mathrm{r}\mathrm{a}\mathrm{d}}{1^{\circ }}\right)=\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}$.

Determination of the equation of angular displacement of pendulum

The relation of displacement of a pendulum is given by,

$x=A\cos \omega t$ ….. (i)

If $\theta$is the angle swept by the pendulum of length l at a particular instant of time $t$, then the arc length can be expressed as the product of angle and the radius$\left(x=l\theta \right)$. So, the equation (i) can be expressed as:

$\begin{array}{rl}cl\theta & =l{\theta }_{max}\cos \omega t\\ \theta & ={\theta }_{max}\cos \omega t\end{array}$

If fis the frequency of the simple pendulum, then the angular frequency of the simple pendulum is given as:

$\omega =2\pi f$

Thus, the angular displacement $\left(\theta \right)$of the pendulum at any time t can be written as:

$\theta ={\theta }_{max}\cos \left(2\pi ft\right)$

On substituting the given values in above equation, you will get:

$\begin{array}{rl}l\theta & =\left(\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}\right)\cos \left(2\pi \left(2.5\mathrm{H}\mathrm{z}\right)t\right)\\ \theta & =\left(\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}\right)\cos \left(5.0\pi t\right)\end{array}$ ….. (ii)

Determination of the position of the pendulum at t = 0.25 s

Using (ii), the position of the pendulum at $t=0.25\mathrm{s}$ is:

$\begin{array}{rl}c{\theta }_{0.25}& =\left(\frac{\pi }{15}\right)\cos (5.0\pi (0.25\mathrm{s}))\\ & =-0.148\mathrm{r}\mathrm{a}\mathrm{d}\\ & \approx -0.15\mathrm{r}\mathrm{a}\mathrm{d}\end{array}$

Therefore, the position of the pendulum at $t=0.25$s is approximately -0.15 rad.

Determination of the position of the pendulum at t = 1.60 s

On substituting $t=1.60\mathrm{s}$ in (ii), you will get:

$\begin{array}{rl}c{\theta }_{1.60}& =\left(\frac{\pi }{15}\right)\cos (\left(5.0\mathrm{H}\mathrm{z}\right)\pi (1.60\mathrm{s}))\\ & =\left(\frac{\pi }{15}\right)\cos 8\pi \\ & =\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}\end{array}$

Therefore, the position of the pendulum at t = 1.60 s is $\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}$.

Determination of the position of the pendulum at t = 500 s

On substituting $t=500\mathrm{s}$ in (ii), you will get:

$\begin{array}{rl}c{\theta }_{500\mathrm{s}}& =\left(\frac{\pi }{15}\right)\cos (\left(5.0\mathrm{H}\mathrm{z}\right)\pi (500\mathrm{s}))\\ & =\left(\frac{\pi }{15}\right)\cos 2500\pi \\ & =\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}\end{array}$

Therefore, the position of the pendulum at $t=500$s is $\frac{\pi }{15}\mathrm{r}\mathrm{a}\mathrm{d}$.