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Expert-verified Found in: Page 472 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # If a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the frequency of its motion? (a) It doubles. (b) It becomes times $\sqrt{2}$ as large. (c) It becomes half as large. (d) It becomes times $\frac{1}{\sqrt{2}}$as large. (e) It remains the same.

Option (d) is the correct answer, i.e it becomes times $\frac{1}{\sqrt{2}}$as larger.

See the step by step solution

## Step 1: Relationship between amplitude and length

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

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

$T=$Period of oscillation

$L=$Length of pendulum

$g=$Gravitational acceleration

## Step 2: Find what will happen when its length is doubled if a simple pendulum oscillates with small amplitude

The period of a simple pendulum is

$T=2\pi \sqrt{\frac{L}{g}}$, and its frequency is $f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{g}{L}}$

Thus, if the length is doubled so$l\text{'}=2l$ , the new frequency is

$f\text{'}=\frac{1}{T\text{'}}=\frac{1}{2\pi }\sqrt{\frac{g}{L\text{'}}}$

$f\text{'}=\frac{1}{2\pi }\sqrt{\frac{g}{2L}}\phantom{\rule{0ex}{0ex}}f\text{'}=\frac{1}{\sqrt{2}}.\frac{1}{2\pi }\sqrt{\frac{g}{L}}\phantom{\rule{0ex}{0ex}}f\text{'}=\frac{1}{\sqrt{2}}.f\phantom{\rule{0ex}{0ex}}$

Hence option (d) is the correct answer for this question. ### Want to see more solutions like these? 