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A simple pendulum oscillates with frequency \(f\). What is its frequency if the entire pendulum accelerates at 0.35 g (a) upward, and (b) downward?

Short Answer

Expert verified

The frequency of the simple pendulum when pendulum is accelerate upwards or downwards is (a) \(1.16f\) and (b) \(0.81f\), respectively.

Step by step solution

01

Concept of frequency for SHM

In this condition, the frequency of simple pendulum is evaluated by using the relation of frequency in simple harmonic motion along with gravitational acceleration and string’s length.

02

Given data

The frequencyis\(f\).

The perturbation in acceleration due to gravity is\(g' = 0.35g\).

03

Calculation of frequency when acceleration is in upward direction 

The revised frequency of a simple pendulum if the entire pendulum is accelerating upwards is calculated as:

\({f_{{\rm{up}}}} = \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{{g + g'}}{l}} \)

Here, l is the length of a simple pendulum.

Substitute the values in the above relation.

\(\begin{aligned}{c}{f_{{\rm{up}}}} &= \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{{g + 0.35g}}{l}} \\{f_{{\rm{up}}}} &= \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{{{\rm{1}}{\rm{.35}}g}}{l}} \\{f_{{\rm{up}}}} &= \sqrt {1.35} f\\{f_{{\rm{up}}}} &= 1.16f\end{aligned}\)

Thus, the required frequency is \(1.16f\).

04

Calculation of frequency when acceleration is in downward direction

The revised frequency of a simple pendulum if the entire pendulum is accelerating downwards is calculated as:

\({f_{{\rm{down}}}} = \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{{g - g'}}{l}} \)

Substitute the values in the above relation.

\(\begin{aligned}{c}{f_{{\rm{down}}}} &= \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{{g - 0.35g}}{l}} \\{f_{{\rm{down}}}} &= \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{{{\rm{0}}{\rm{.65}}g}}{l}} \\{f_{{\rm{down}}}} &= \sqrt {0.65} f\\{f_{{\rm{down}}}} &= 0.81f\end{aligned}\)

Thus, the required frequency is\(0.81f\).

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