Solve the problem.
-The position (in centimeters) of an object oscillating up and down at the end of a spring is given by $s = A \sin \left( \sqrt { \frac { k } { m } } t \right)$ at time $t$ (in seconds). The value of A is the amplitude of the motion, $k$ is a measure of the stiffness of the spring, and $m$ is the mass of the object. How fast is the object moving when it is moving fastest?
A) $A \sqrt { \frac { \mathrm { m } } { \mathrm { k } } } \mathrm { cm } / \mathrm { sec }$
B) $\sqrt { \frac { \mathrm { k } } { \mathrm { m } } } \mathrm { cm } / \mathrm { sec }$
C) $\mathrm { A } \mathrm { cm } / \mathrm { sec }$
D) $A \sqrt { \frac { \mathrm { k } } { \mathrm { m } } } \mathrm { cm } / \mathrm { sec }$

Question

Quizplus · Accepted Answer

The velocity of the object can be found by taking the derivative of the displacement equation with respect to time: 
$$v=\frac{ds}{dt}=A\sqrt{\frac{k}{m}}\cos\left(\sqrt{\frac{k}{m}}tight)$$
The object is moving fastest when the cosine term is equal to 1 (i.e. at the maximum of the cosine function): 
$$\cos\left(\sqrt{\frac{k}{m}}tight) = 1$$
This occurs at $t=0$ and any time that satisfies the equation $\sqrt{\frac{k}{m}}t = n\pi,$ where $n$ is an integer. Thus, the maximum velocity of the object is: 
$$v_	ext{max} = A\sqrt{\frac{k}{m}}\cos(0) = A\sqrt{\frac{k}{m}}$$
Therefore, the answer is choice D, $A\sqrt{\frac{k}{m}}$ cm/sec.

Solve the Problem s=Asin⁡(kmt)s = A \sin \left( \sqrt { \frac { k } { m } } t \right)s=Asin(mk​​t)

Solve the Problem $s = A \sin \left( \sqrt { \frac { k } { m } } t \right)$