JEE Main 2014 · 19 Apr · Shift 1 · Physics Q14

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PhysicsSingle correct+4 / -1Question 14 of 90

A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

4 cm/s

6 cm/s

12 cm/s

16 cm/s

Answer and worked solution

Official answer: C

$x=A\sin\omega t$ reaches $\frac A2$ at $\omega t=\frac\pi6$ , i.e. $t=\frac T{12}=\frac1{24}$ s.
$\bar v=\frac{0.5\text{ cm}}{1/24\text{ s}}=12\text{ cm/s}.$

Topics:Simple harmonic motion

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 14