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JEE MAIN 2025
02-04-2025 SHIFT-1
Question
A particle is subjected to two simple harmonic motions as:
${x_1} = \sqrt 7 \sin 5tcm$ and ${x_2} = 2\sqrt 7 \sin \left( {5t + \frac{\pi }{3}} \right)cm$
where x is displacement and t is time in seconds.
The maximum acceleration of the particle is $x \times {10^{-2}}m{s^{-2}}$. The value of x is :
Select the correct option:
A
$5\sqrt 7 $
B
$25\sqrt 7 $
C
175
D
125
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Physics
Easy
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