SIMPLE HARMONIC MOTION_{S2}_JEE_MAIN_12-01-2019 - Competishun

JEE MAIN 2019

12-01-2019 S2

Question

A simple harmonic motion is represented by: $$ y=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) c m $$ The amplitude and time period of the motion are:

Select the correct option:

$10 \mathrm{~cm}, \frac{2}{3} \mathrm{~s}$

$5 \mathrm{~cm}, \frac{2}{3} \mathrm{~s}$

$10 \mathrm{~cm}, \frac{3}{2} \mathrm{~s}$

$5 \mathrm{~cm}, \frac{3}{2} \mathrm{~s}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Easy

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