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Oscillations

Question
CBSEENPH11026308
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A mass of 1 kg is suspended from a spring of force constant 400 N, executing SHM total energy of the body is 2J, then maximum acceleration of the spring will be

  • 4 m/s2

  • 40 m/s2

  • 200 m/s2

  • 400 m/s2

Solution

B.

40 m/s2

Oscillations due to spring

           T = 2π mR

Where T is the time period

          k is spring constant

          m is mass attached to spring

         T = 2πmR

              = 2π 1400

              = 2π20

    Frequency n = 202 π

     ω = 2π n

         = 2π × 202π

       ω = 20

  Total energy

           E = 12mω2 r2

            2 = 12m ω2r2

            r = 110 m

   Acceleration = ω2 r

                      = 400 × 110

                       = 40 m/s2

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