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Eduardwww
2 months 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:
serg [3.5K]2 months ago
4 0

Answer:

The amplitude A(5) = 1 \ m

Explanation:

The problem states 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 frame evaluated is  t = 5.0 s

In general, the displacement (Amplitude) of damped harmonic motion can be mathematically defined as

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

Now replacing the values

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

         A(5) = 1 \ m

       

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