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Found in: Page 436

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# What is the maximum acceleration of a platform that oscillates at amplitude 2.20 cm and frequency 6.60 Hz?

Maximum value of acceleration is 37.8 m/s2.

See the step by step solution

## Step 1: Stating the given data

1. Amplitude of platform oscillation, $\mathrm{x}=2.2\mathrm{cm}\mathrm{or}2.2×{10}^{-2}{\mathrm{m}}^{2}$.
2. Frequency of platform oscillation, $\mathrm{f}=6.60\mathrm{Hz}$.

## Step 2: Understanding the concept of acceleration

In a simple harmonic motion, the body undergoes acceleration for amplitude. Maximum acceleration occurs when the object is at end of its path. At those points, the force acting on the object is also maximum.

Formula:

The acceleration of a body in simple harmonic motion is directly proportional to the displacement, given by

${\mathrm{a}}_{\mathrm{m}}={\mathrm{\omega }}^{2}\mathrm{x}$ (i)

## Step 3: Calculation of acceleration of the body in motion

To find maximum acceleration, the angular frequency can be shown as follows:

$\begin{array}{rcl}\mathrm{\omega }& =& 2\mathrm{\pi f}\\ & =& 2\mathrm{\pi }×6.60\left(\because \mathrm{f}=6.60\mathrm{Hz}\right)\\ & =& 41.448\mathrm{rad}/\mathrm{sec}\end{array}$

Now using equation (i) and the given & derived values, the maximum acceleration is as follows:

$\begin{array}{rcl}{\mathrm{a}}_{\mathrm{m}}& =& 41.{469}^{2}×2.2×{10}^{-2}\\ & =& 37.795\mathrm{m}/{\mathrm{s}}^{2}\\ & \simeq & 37.8\mathrm{m}/{\mathrm{s}}^{2}\end{array}$

Hence, the value of maximum acceleration is $37.8\mathrm{m}/{\mathrm{s}}^{2}$.