Question

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to rest. It is then pulled down 6.70 cm and released.
What is the time constant if the ball's amplitude has decreased to 2.20 cm after 30.0 oscillations?

Answer 1

Answer:

The time constant is $\tau = 17.5 \ s$    

Explanation:

From the question we are told that

   The spring constant is  $k = 11.5 \ N/m$

   The mass  of the ball is  $m_b = 490 \ g = 0.49 \ kg$

   The amplitude of the  oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

    The amplitude after time t is  $x_t = 2.20 cm = 0.022 \ m$

    The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

       $t = N * T$                                            Here  T is the period of oscillation

         $t = N * 2\pi \sqrt{\frac{m}{k} }$

=>     $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=>     $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

         $x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping  constant so

 at  $t = 38.88 \ s$ ,  $x_t = 2.20 cm = 0.022 \ m$

So  

     $0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=>  $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=>  $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$    

=>   $a = 0.028$

Generally the time constant is mathematically represented as

    $\tau = \frac{m}{a}$      

=> $\tau = \frac{0.490}{ 0.028}$    

=> $\tau = 17.5 \ s$