Question

A 350 g mass on a 45-cm-long string is released at an angle of 4.5° from vertical. It has a damping constant of 0.010 kg/s. After 25 s, how many oscillations has it completed?

Answer 1

Answer:

The value is  $n = 18.5 \ oscillations$

Explanation:

From the question we are told that

   The mass is  $m = 350 \ g = 0.350 \ kg$

   The length is  $L = 45 \ cm = 0.45 \ m$

    The angle is  $\theta = 4.5^o$

    The damping constant is  $b = 0.010 \ kg/s$

    The time taken is $t = 25 \ s$

Generally the angular frequency of this damped oscillation is mathematically evaluated as

     $w = \sqrt{ \frac{ g}{L} + \frac{b^2}{4m^2} }$  

=>   $w = \sqrt{ \frac{9.80 }{ 0.45} + \frac{0.010 ^2}{4* 0.350^2} }$  

=>   $w = 4.667 \ s^{-1}$

Generally the period of the oscillation is mathematically represented as

      $T = \frac{2 \pi }{w}$  

=>  $T = \frac{2 * 3.142 }{ 4.667 }$

=>  $T = 1.35 \ s$

Generally the number of oscillation is mathematically represented as

        $n = \frac{t}{T}$

=>     $n = \frac{25}{1.35}$

=>     $n = 18.5 \ oscillations$