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JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Elasticity & SHM

The displacement of a particle executing SHM is given by

x = 10 sin (ω t + \frac{π}{3})

m. The time period of motion is 3.14 s . The velocity of the particle at

t = 0

is ________

m/s

.

Answer: 10

Solution

Given

T = 3.14 s

, angular frequency

ω = \frac{2 π}{T} = \frac{2 π}{π} = 2 rad/s

(taking

π \approx 3.14

).

x = 10 sin (ω t + \frac{π}{3}) ⟹ v = \frac{d x}{d t} = 10 ω cos (ω t + \frac{π}{3})

At

t = 0

:

v = 10 (2) cos (\frac{π}{3}) = 20 \times \frac{1}{2} = 10 m/s

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Integer answer

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