Solved: PHYSICS Question for NEET | Sushrut

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NEET PHYSICSEasy

Two simple harmonic motions of angular frequency $100\text{ rad s}^{-1}$ and $1000\text{ rad s}^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration will be:

A

$1:10$

B

$1:10^2$

C

$1:10^3$

D

$1:10^4$

Step-by-Step Solution

Identify the Formula: The maximum acceleration ( $a_{max}$ ) of a particle executing simple harmonic motion is given by the formula $a_{max} = \omega^2 A$ , where $\omega$ is the angular frequency and $A$ is the displacement amplitude .
Analyze Given Data:

Angular frequency of first SHM, $\omega_1 = 100\text{ rad s}^{-1}$ .
Angular frequency of second SHM, $\omega_2 = 1000\text{ rad s}^{-1}$ .
The amplitude ( $A$ ) is the same for both motions.

Calculate Ratio: $\frac{a_{1}}{a_{2}} = \frac{\omega_1^2 A}{\omega_2^2 A} = \left(\frac{\omega_1}{\omega_2}\right)^2$ $\frac{a_{1}}{a_{2}} = \left(\frac{100}{1000}\right)^2 = \left(\frac{1}{10}\right)^2 = \frac{1}{100}$

The ratio is $1:10^2$ .

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