Question

As you enter a fine restaurant, you realize that you have accidentally brought a small electronic timer from home instead of your cell phone. In frustration, you drop the timer into a side pocket of your suit coat, not realizing that the timer is operating. The arm of your chair presses the light cloth of your coat against your body at one spot. Fabric with a length L hangs freely below that spot, with the timer at the bottom. At one point during your dinner, the timer goes off and a buzzer and a vibrator turn on and off with a frequency of 1.50 Hz. It makes the hanging part of your coat swing back and forth with remarkably large amplitude, drawing everyone’s attention. Find the value of L.

Short answer

The value of L is 11 cm.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The lengthoffabric is, L.
• The frequency of the oscillation is, f = 1.50 Hz.

Significance of the simple pendulum

A simple pendulum of length L can be modeled to move in simple harmonic motion for small angular displacements from the vertical. Its period is

$\mathit{T}\mathbf{=}\mathbf{2}\mathit{\pi }\sqrt{\frac{\mathbf{L}}{\mathbf{g}}}$

Determination of the length L

Let the time period is T;

Byusingconceptandformulafromstep(1),weget

$T=2\pi \sqrt{\frac{L}{g}}$

But$f=\frac{1}{T}$

Therefore, it becomes

$$\begin{gathered} f=\frac{1}{2\pi }\sqrt{\frac{g}{L}} \\ L=\frac{g}{f^2\times {\left(2\pi \right)}^2} \end{gathered}$$

Substitute all the value in the above equation,

$$\begin{gathered} L=\frac{10 m{s}^{-2}}{{\left(1.5 \text{Hz}\right)}^2\times 4\times {\pi }^2} \\ L=0.11 \text{m} \\ L=11 \text{cm} \end{gathered}$$

Hence the length is 11 cm.