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AP EAMCET20226 Jul 2022Morning ShiftPhysicsOscillationsActual

A point mass of 400 ~g executes S.H.M. under a force F =- (10 Nm ⁻¹ ) x . If it crosses the centre of oscillation with a speed of 10 ~ms ⁻¹ , the amplitude of motion is

Options

  1. A2 ~m
  2. B4 ~m
  3. C0.4 ~m
  4. D0.5 ~m

Correct answer

B. 4 ~m

Step-by-step solution

We have, F =-10 x Comparing it with F =- m ^2 x , we get m ^2=10 = 10 ~m = 10 0.4 =5 rad / sec Now, V ₀= A aligned & 10=5 ~A & A =2 ~m aligned

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