JEE Main 2026 · 6 Apr · Shift 1 · Physics Q37

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PhysicsSingle correct+4 / -1Question 37 of 75

A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec. The time required by it to move from $x=A$ to $x=\dfrac A{\sqrt2}$ is ________ sec.

1/4

5/4

5/8

3/8

Answer and worked solution

Official answer: C

Phase. With $x=A\cos\omega t$ , $x=\tfrac A{\sqrt2}$ when $\omega t=\tfrac\pi4$ .

Time. $t=\dfrac T8=\dfrac58$ s.

(C)

Topics:Simple harmonic motion

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JEE Main 2026 · 6 Apr · Shift 1, Physics question 37