JEE Main 2022 · 29 Jun · Shift 1 · Physics Q54

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PhysicsInteger+4 / -1Question 54 of 90

A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is $\sqrt x$ cm. The value of $x$ is ______.

Your answer

Answer and worked solution

Official answer: 700

Speed at 5 cm. $v=\omega\sqrt{100-25}=\omega\sqrt{75}$ ; tripled to $3\omega\sqrt{75}$ .

New amplitude. $A'^2=25+9\times75=700$ .

Answer: 700

Topics:Energy in SHM Simple harmonic motion

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JEE Main 2022 · 29 Jun · Shift 1, Physics question 54