Question

The position of an object in simple harmonic motion is given by x= (6.88 cm) cos [(2 pie/0.663 s)t]

(a) What is the object's speed at 0.828 s?

cm/s

(b) What is the object's maximum speed?

cm/s

(c) What is the object's speed when -6.88 cm?

cm/s

Answer #1

(a) Expression for the position is given by -

x = (6.88 cm) cos [(2 pie/0.663 s)t]

Speed of the object, v = dx/dt = [-(6.88 x 2x pie)/0.633 cm] sin [(2 pie/0.663 s)t] = -(68.3 cm/s) sin [(2 pie/0.663 s)t]

At t = 0.828 s -

v = -(68.3 cm/s) sin [(2 pie/0.663 s)*0.828s] = -68.29 cm/s

(b) Maximum speed of the object = 68.3 cm/s.

(c) When displacement = -6.88 cm, find the time t -

=> -6.88 cm = (6.88 cm) cos [(2 pie/0.663 s)t]

=> cos [(2 pie/0.663 s)t] = -1

=> (2 pie/0.663 s)t = pie

=> t = 0.663/2 s

Now put this value of t and find the speed of the object.

Means -

v = -(68.3 cm/s) sin [(2 pie/0.663 s)*0.663/2s] = -(68.3 cm/s) sin [ pie] = 0

So, speed of the object = 0

Hope, you understand the solution!

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