Chapter 12, Problem 45QAP

To determine

(a)

The position function of the object.

Answer to Problem 45QAP

The position of the object is $x(t)=A\cos \left(\sqrt{\frac{k}{M}}t\right)$

Explanation of Solution

Introduction:

The displacement of the spring performing motion is given by,

  $x(t)=A\cos (\omega t)$

Where,

  $x(t)$ = position of object at time t A = amplitude

t = time

  $\omega$ =angular velocity

The displacement of the spring performing motion is given by,

  $x(t)=A\cos (\omega t)$

But,

  $\begin{array}{l}\omega =\sqrt{\frac{k}{M}}\\ \therefore x(t)=A\cos \left(\sqrt{\frac{k}{M}}t\right)\end{array}$

Conclusion:

The position of the object is $x(t)=A\cos \left(\sqrt{\frac{k}{M}}t\right)$

To determine

(b)

The velocity of the object at $t=\frac{7T}{6}$.

Answer to Problem 45QAP

The velocity of the object at $t=\frac{7T}{6}$ is, $v(t)=-A\sqrt{\frac{k}{M}}\sin \left(\sqrt{\frac{k}{M}}\times \frac{7T}{6}\right)$

Explanation of Solution

Introduction:

The velocity of the particle performing simple harmonic motion,

  $v(t)=-A\omega \sin (\omega t)$

Where,

  $v(t)$ = velocity of object at time t A = amplitude

t = time

  $\omega$ =angular velocity

Calculation:

The velocity of the particle performing simple harmonic motion,

  $\begin{array}{l}v(t)=-A\omega \sin (\omega t)\\ v(t)=-A\sqrt{\frac{k}{M}}\sin \left(\sqrt{\frac{k}{M}}\times \frac{7T}{6}\right)\end{array}$

Conclusion:

The velocity of the object at $t=\frac{7T}{6}$ is, $v(t)=-A\sqrt{\frac{k}{M}}\sin \left(\sqrt{\frac{k}{M}}\times \frac{7T}{6}\right)$

To determine

(c)

The acceleration of the object at

  $t=\frac{T}{4}$.

Answer to Problem 45QAP

The acceleration of the object at $t=\frac{T}{4}$ is, $a(t)=0$

Explanation of Solution

Introduction:

The acceleration of the particle performing simple harmonic motion,

  $a(t)=-A{\omega }^2\cos (\omega t)$

Where,

  $a(t)$ = acceleration of object at time t A = amplitude

t = time

  $\omega$ =angular velocity

Calculation:

The acceleration of the particle performing simple harmonic motion,

  $\begin{array}{l}a(t)=-A{\omega }^2\cos (\omega t)\\ a(t)=-A{\omega }^2\cos \left(\frac{2\pi }{T}\times \frac{T}{4}\right)\\ a(t)=-A{\omega }^2\cos \left(\frac{\pi }{2}\right)\\ a(t)=0\end{array}$

Conclusion:

The acceleration of the object at $t=\frac{T}{4}$ is,

  $a(t)=0$