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JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Elasticity & SHM

A particle of mass 0.50 kg executes simple harmonic motion under force

F = - 50 (Nm^{- 1}) x

. The time period of oscillation is

\frac{x}{35} s

. The value of

x

is ................
(Given

π = \frac{22}{7}

)

Answer: 22

Solution

F = - 50 x

N on

m = 0.5

kg gives

a = \frac{F}{m} = - 100 x ⟹ ω^{2} = 100 ⟹ ω = 10 rad/s

T = \frac{2 π}{ω} = \frac{2}{10} \cdot \frac{22}{7} = \frac{22}{35} s

Comparing with

T = \frac{x}{35}

s:

x = 22

.

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Integer answer

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