Question

A mass on a spring bounces up and down in simple harmonic motion, modeled by the function s ( t ) = 3 sin t(t)=3 sin t where s is measured in centimeters and t t is measured in seconds. Find the rate at which the spring is oscillating at t = 9 t=9 s. Round your answer to four decimal places.

Answer 1

Answer:

Rate at which spring is oscillating 2.9631

Step-by-step explanation:

The governing equation for the position of the spring = 3 sin t

the change in position of the spring with respect to time is =$\frac{ds}{dt}=3cost$

given time t = 9 seconds,

To find the rate at which the spring is oscillating at the time, we insert the time 9 seconds into the the slot for t

$\frac{ds}{dt}=3cos 9= 2.9631$

rate at which it is oscillating = 2.9631

Answer 2

The rate at which the spring is oscillating at t = 9 s is -2.7333 cm/s

Since the simple harmonic motion, modeled by the function s(t) = 3 sin t where s is measured in centimeters and t is measured in seconds.

The rate of oscillation

The rate at which the spring is oscillating is the rate of change of s(t) with respect to time.

Since s(t) = 3sint

So, ds(t)/dt = d(3sint)/dt

ds(t)/dt = 3dsint/dt

ds(t)/dt = 3cost

The rate of oscillation at t = 9 s

Now, to find the rate at which the spring is oscillating at t = 9, we substitute t = 9 into ds(t)/dt.

So, ds(t)/dt = 3cost

ds(t)/dt = 3cos9

ds(t)/dt = 3 × -0.9111

ds(t)/dt = -2.7333 cm/s

So, the rate at which the spring is oscillating at t = 9 s is -2.7333 cm/s

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