Chapter 5.6, Problem 36E

(a)

To determine

To find: The equation of theoscillation of the spring.

Answer to Problem 36E

The equation of the motion of the massis $f\left(t\right)=5\cos \left(17.32t\right)$

Explanation of Solution

Given Information:

The mass is $10\text{g}\text{or}0.01\text{kg}$ , the stiffness of spring is $3$ , and the amplitude of the motion is $5\text{cm}$ .

Concept used:

The standard form of the simple harmonic motion when the displacement is maximum at $t=0$ is given by,

  $y=A\cos \omega t$

The angular velocity is given by:

  $\omega =\frac{2\pi }{T}$

The function of the simple harmonic motion in time period term is given by,

  $y=A\cos \left(\frac{2\pi t}{T}\right)$

The equation of the motion of the massis calculated as,

  $\begin{array}{c}f\left(t\right)=a\cos \left(\sqrt{\frac{k}{m}}t\right)\\ =5\cos \left(\sqrt{\frac{3}{0.01}}t\right)\\ =5\cos \left(17.32t\right)\end{array}$

Thus, the equation of the motion of the massis $f\left(t\right)=5\cos \left(17.32t\right)$ .

(b)

To determine

To find: The general formula for the frequency.

Answer to Problem 36E

The general formula for the frequency is $N=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$ .

Explanation of Solution

Given Information:

The mass is $10\text{g}\text{or}0.01\text{kg}$ , the stiffness of spring is $3$ , and the amplitude of the motion is $5\text{cm}$

Concept used:

The standard form of the simple harmonic motion when the displacement is maximum at $t=0$ is given by,

  $y=A\cos \omega t$

The angular velocity is given by:

  $\omega =\frac{2\pi }{T}$

The function of the simple harmonic motion in time period term is given by,

  $y=A\cos \left(\frac{2\pi t}{T}\right)$

The equation of the motion of the mass is calculated as,

  $f\left(t\right)=a\cos \left(\sqrt{\frac{k}{m}}t\right)$

The standard form of the simple harmonic motion when the displacement is maximum at $t=0$ is given by,

  $f\left(t\right)=a\cos \left(2\pi Nt\right)$

Compare both the equation to obtain,

  $\begin{array}{l}2\pi N=\sqrt{\frac{k}{m}}\\ N=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\end{array}$

Thus, the general formula for the frequencyis $N=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$ .