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 $\mathrm { m }$ is the mass of the object. How fast is the object accelerating when it is accelerating the fastest?
A) $\mathrm { A } ^ { 2 } \mathrm {~cm} / \mathrm { sec } ^ { 2 }$
B) $A \sqrt { \frac { \mathrm { k } } { \mathrm { m } } } \mathrm { cm } / \mathrm { sec } ^ { 2 }$
C) $\mathrm { A } \mathrm { cm } / \mathrm { sec } ^ { 2 }$
D) $\frac { \mathrm { Ak } } { \mathrm { m } } \mathrm { cm } / \mathrm { sec } ^ { 2 }$

Question

Quizplus · Accepted Answer

The acceleration of the object is the second derivative of the position with respect to time. Differentiating $s = A \sin \left( \sqrt { \frac { k } { m } } t \right)$ twice with respect to $t$ gives acceleration $a = -A \left( \frac{k}{m} \right) \sin \left( \sqrt { \frac { k } { m } } t \right)$. The maximum value of $\sin$ function is 1, so the maximum acceleration is $\frac { \mathrm { Ak } } { \mathrm { m } } \mathrm { cm } / \mathrm { sec } ^ { 2 }$.

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

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