Question

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?

Answer 1

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

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

         $A(5) = 1 \ m$