Question

What is the frequency of a simple pendulum $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$long (a) in a room, (b) in an elevator accelerating upward at a rate of role="math" localid="1657259780987" $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, and (c) in free fall?

Short answer

1. The frequency of the pendulum in a room is$0.35\mathrm{Hz}$.
2. The frequency of the pendulum in an elevator is$0.39\mathrm{Hz}$.
3. The frequency of pendulum in free fall is $0\mathrm{Hz}$.

Step-by-step solution

The given data

Length of the pendulum$L=20\mathrm{m}$

Understanding the concept of frequency

The angular frequencyis related to the periodand frequencyof the motion by,

$\begin{array}{l}\mathbf{\omega }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{T}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{\pi f}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}\end{array}$

Using the relation between frequency and acceleration and then considering acceleration in various situations we can find frequencies in the room, in the elevator with acceleration, and in the free fall case.

Formula:

The angular frequency of a body in oscillation,

$\omega =2\pi for\omega =\frac{2\pi }{T}or\omega =\sqrt{\left(\frac{mgL}{l}\right)}$ (i)

The period of oscillation of a pendulum,$T=2\pi \sqrt{\frac{L}{g}}$ (ii)

The frequency of a body, $f=1/T$ (iii)

a) Calculation of frequency in the room

Using equations (ii) and (iii), we can get the frequency in the room as:

$$\begin{gathered} f\left(r\right)=\frac{1}{2\times 3.14}\sqrt{\left(\frac{9.8\mathrm{m}/{\mathrm{s}}^2}{2\mathrm{m}}\right)} \\ =0.3523\mathrm{Hz} \\ \approx 0.35\mathrm{Hz} \end{gathered}$$

Hence, the required value of frequency is $0.35\mathrm{Hz}$.

b) Calculation of frequency in the elevator

As the elevator is moving upward, acceleration a is positive and we have to add it with g. Hence, again using the equations (ii) and (iii), we get the frequency in the elevator as:

$$\begin{gathered} f\left(e\right)=\frac{1}{2\mathrm{\pi }}\sqrt{\left(\frac{g+a}{L}\right)} \\ =\frac{1}{2\times 3.14}\sqrt{\left(\frac{9.8\mathrm{m}/{\mathrm{s}}^2+2\mathrm{m}/{\mathrm{s}}^2}{2\mathrm{m}}\right)} \\ =0.3853\mathrm{Hz} \\ \approx 0.39\mathrm{Hz} \end{gathered}$$

Hence, the required value of frequency is $0.39\mathrm{Hz}$.

c) Calculation of frequency in the free-fall

As the pendulum is in free fall, acceleration is negative and is equal to g. Hence, using the equations (ii) and (iii), we get the frequency in the free-fall case as:

$$\begin{gathered} f\left(f\right)=\frac{1}{2\mathrm{\pi }}\sqrt{\left(\frac{g-g}{L}\right)} \\ =0\mathrm{Hz} \end{gathered}$$

Hence, the required value of frequency is $0\mathrm{Hz}$.