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Mila [183]
2 years ago
13

A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the

mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?
Physics
1 answer:
netineya [11]2 years ago
4 0

Answer:

The amplitude A(5) = 1 \ m

Explanation:

From the question we are told that

     The  spring constant is  k =  100 \ N/m

      The  damping constant is  b = 8.0 *10^{-3} \  kg \cdot m/s

       The mass is  m =  0.050 \ kg

       The  maximum displacement is A_o  =  1.5 \ m  \ at  t = 0

       The  time  considered is  t =  5.0 s

Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as

           A(t) =  A_o *   e ^{ - \frac{b *  t}{2 *  m} }

substituting values

         A(5) =  1.5  *   e ^{ - \frac{ 8.0 *10^{-3} *  5}{2 *  0.050} }

         A(5) = 1 \ m

       

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