Question

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is_______ cm. [Assume that the mass of the pendulum remains same]}

• A. 7.5
• B. 120
• C. 0.75
• D. 15

Hint:

To double the number of oscillations in the same time (double the frequency), you must reduce the length of the pendulum to one-fourth of its original value.

Answer 1

The Correct Option is A

To solve this problem, we need to find the length of the string required for a simple pendulum to perform 40 oscillations in the same time it took to perform 20 oscillations. We start by understanding the relationship between the length of the pendulum and its period.

The time period \(T\) of a simple pendulum is given by the formula: \(T = 2\pi \sqrt{\frac{L}{g}}\) where:

• \(T\) is the period of one complete oscillation,
• \(L\) is the length of the pendulum,
• \(g\) is the acceleration due to gravity (approx. \(9.8 \, \text{m/s}^2\)).

 

The number of oscillations \(n\) in a given time \(t\) is related to the period by: \(n = \frac{t}{T}\)

Let's solve for the initial pendulum length settings. For the first case, with 20 oscillations in 10 seconds:

• Number of oscillations \(n_1 = 20\)
• Time \(t = 10 \, \text{s}\)
• Length \(L_1 = 30 \, \text{cm} = 0.30 \, \text{m}\)

The time period \(T_1\) is then: \(T_1 = \frac{t}{n_1} = \frac{10}{20} = 0.5 \, \text{s}\)

Using the formula for the period: \(0.5 = 2\pi \sqrt{\frac{0.30}{g}}\)

Now, for the second case where we want 40 oscillations in 10 seconds:

• Number of oscillations \(n_2 = 40\)

The time period \(T_2\) is then: \(T_2 = \frac{t}{n_2} = \frac{10}{40} = 0.25 \, \text{s}\)

Using the period formula again: \(0.25 = 2\pi \sqrt{\frac{L_2}{g}}\)

Solving for \(L_2\) gives: \(\frac{0.25}{2\pi} = \sqrt{\frac{L_2}{g}}\) 
\(L_2 = g \left(\frac{0.25}{2\pi}\right)^2\)

We already know from the first setting: \(\frac{0.5}{2\pi} = \sqrt{\frac{0.30}{g}}\)

Then using ratio: \(\left(\frac{\frac{0.25}{2\pi}}{\frac{0.5}{2\pi}}\right)^2 = \frac{L_2}{0.30}\) 
\(\left(\frac{0.25}{0.5}\right)^2 = \frac{L_2}{0.30}\) 
\(L_2 = 0.30 \times \left(\frac{1}{2}\right)^2 = 0.30 \times \frac{1}{4} = 0.075 \, \text{m} = 7.5 \, \text{cm}\)

Thus, the length of the string required for the pendulum to perform 40 oscillations in 10 seconds is 7.5 cm.