A 2.50 kg frictionless block is attached to an ideal spring with force constant

Question

A 2.50 kg frictionless block is attached to an ideal spring with force constant 314 N/m is undergoing simple harmonic motion. When the block has displacement 0.270 m , it is moving in the negative x-direction with a speed of 3.82 m/s .

Find the amplitude of the motion.

Express your answer with the appropriate units.

Part B

Find the magnitude of the block's maximum acceleration.

Express your answer with the appropriate units.

Find the magnitude of the maximum force the spring exerts on the block.

Express your answer with the appropriate units.

Part B

Find the magnitude of the block's maximum acceleration.

Express your answer with the appropriate units.

Find the magnitude of the maximum force the spring exerts on the block.

Express your answer with the appropriate units.

Explanation / Answer

A)
w = sqrt (k/m)
= sqrt (314/2.5)
= 11.21 rad/s

x = A* cos (w*t)
v = -A*w*sin (w*t)

divide 2nd equation by 1st one:
v/x = -w*tan (w*t)
3.82/0.27 = - 11.21 * tan (11.21*t)
tan (11.21*t) = -1.26
11.21 * t = 0.9 rad
t = 0.08 s

x = A* cos (w*t)
-0.27 = A* cos (11.21 * 0.08)
A = 0.27 m approx
Answer: 0.27 m

B)
Max a = A^2 *w = 0.27^2 * 11.21 = 0.82 rad/s^2

C)
max F= A*K = 0.27 * 314 = 84.78 m

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