Question

Question: Suppose that a simple pendulum consists of a small 60.0 g bob at the end of a cord of negligible mass. If the angle between the cord and the vertical is given by,

$\mathit{\theta }\mathbf{=}\left(0.0800\text{rad}\right)\mathbf{\text{cos}}\left[\left(4.43\text{rad}/\text{s}\right)t+\right]$,

1. What is the pendulum’s length?
2. What is its maximum kinetic energy?

Short answer

Answer

1. The length of the pendulum is 0.499m

The maximum kinetic energy of the pendulum is$9.40\times {10}^{-4}\text{J}$

Step-by-step solution

Identification of given data

1. The mass of the small bob is $m=60.0\text{g}=60.0\times {10}^{-3}\text{kg}$
2. The angle between the cord and vertical is $\theta =\left(0.0800\text{rad}\right)\text{cos}\left[\left(4.43\text{rad}/\text{s}\right)t+\right]$

Understanding the concept

The oscillations of the simple pendulum can be defined by the equation of simple harmonic motion. The simple harmonic motion is the motion in which the acceleration of the oscillating object is directly proportional to the displacement. The force caused by the acceleration is called restoring force. This restoring force is always directed towards the mean position. A simple harmonic oscillator will have maximum kinetic energy at its mean position because, at the mean position, the velocity of the oscillation is maximum. The maximum velocity of oscillation is equal to the product of amplitude and angular velocity of the pendulum.

Compare the given equation with the equation of displacement of theparticle in simple harmonic motion. Use the formula for maximum velocity to find the maximum kinetic energy.

Formulae:

$x\left(t\right)=x_m\cos \left(\omega t+\right)$

Here, x(t) is the displacement of the pendulum, x_m is maximum displacement or amplitude of oscillations, w is angular frequency of oscillation,t is time, and$\varphi$ is phase difference.

$v_m=x_m\omega$

Here, x_m is maximum velocity of oscillation,w is maximum displacement or amplitude of oscillations, is angular frequency of oscillation,

$K.E_m=\frac{1}{2}m{v}_m^2$

Here, $K.E_m$is maximum kinetic energy, m is mass of the pendulum, $v_m^{}$is maximum velocity of oscillation.

(a) Determining the pendulum’s length

The equation of displacement oftheparticle in simple harmonic motion is

$x\left(t\right)=x_m\cos \left(\omega t+\right)$ …(i)

The angle between the cord and vertical is

$\theta =\left(0.0800\text{rad}\right)\text{cos}\left[\left(4.43\text{rad}/\text{s}\right)\text{t}+\right]$ …(ii)

By comparing equation (i) with equation (ii) as

$$\begin{gathered} \text{x}\left(\text{t}\right)=\theta \\ {\text{x}}_m=0.0800\text{rad} \\ \omega =4.43\text{rad}/\text{s} \end{gathered}$$

The expression for the angular frequency of the pendulum is

$$\begin{gathered} \omega =\sqrt{\frac{g}{L}} \\ {\omega }^2=\frac{g}{L} \\ L=\frac{g}{{\omega }^2} \\ =\frac{9.8\text{m}/{\text{s}}^2}{{\left(4.43\text{rad}/\text{s}\right)}^2} \\ =0.499\text{m} \end{gathered}$$

Therefore, the length is 0.499

(b) Determining the maximum kinetic energy of the pendulum

The expression of velocity amplitude is

${\text{v}}_m={\text{x}}_m\omega$ …(iii)

The equation for arc length is

$$\begin{gathered} \text{Angle}=\frac{\text{arclength}}{\text{radius}} \\ {\theta }_m=\frac{{\text{x}}_m}{\text{L}} \\ {\text{x}}_m={\theta }_m\text{L} \end{gathered}$$

Equation (iii) becomes,

$$\begin{gathered} v_m={\theta }_{m}L\omega \\ =0.0800\text{rad}\times 0.499\text{m}\times 4.43\text{rad/s} \\ =0.1768\text{m/s} \end{gathered}$$

The kinetic energy of the pendulum is

$$\begin{gathered} K.E_m=\frac{1}{2}m{v}_m^2 \\ =\frac{1}{2}\times 60.0\times {10}^{-3}\text{kg}\times {\left(0.1768\text{m/s}\right)}^2 \\ =9.40\times {10}^{-4}\text{J} \end{gathered}$$

Therefore, the kinetic energy of the pendulum is$9.40\times {10}^{-4}\text{J}$ .