A 0.580-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.580-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.2 cm. (Assume the position of the object is at the origin at

t = 0.)

(d) Calculate the value of its acceleration when the object is 9.20 cm from the equilibrium position.

hint; How can you determine the force that acts on the object at this position? cm/s2

(e) Calculate the time interval required for the object to move from x = 0 to x = 3.20 cm.

Hint:
Can you use your equation for x(t) to determine the time for the object to first reach 3.20 cm? s

Explanation / Answer

here ,

d) let the acceleration is a

using second law of motion

m * a = k * x

0.580 * a = 8 * 0.092

a = 1.27 m/s^2 = 127 cm/s^2

the acceleration at this position is 127 cm/s^2

e)

as w = sqrt(k/m)

w = sqrt(8/0.580) = 3.714 rad/s

x = 11.2 * sin(3.714 * t )

Now, for x = 3.20

3.20 = 11.2 * sin(3.714 * t )

solving for t

t = 0.078 s

the time taken to reach 3.2 cm is 0.078 s

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