An ideal spring is hung from the ceiling. A 2.50 kg block is attached to the end

Question

An ideal spring is hung from the ceiling. A 2.50 kg block is attached to the end of the spring. When the block is stationary (in equilibrium) it is found that the spring stretches 25.0 cm from its original length.

The same spring is then placed horizontally, on a frictionless surface. The left end of the spring is fixed to a wall and the 2.50 kg block is attached to the right end. The block is pulled to the right, so as to stretch the spring by 25.0 cm from its unstrained length. The block is released, starting from rest, and simple harmonic motion results.

Find:

a) The spring constant of the spring.

b) The period of the resulting simple harmonic motion.

c) The acceleration (both magnitude and direction) of the block at a time t = 0.500 s after
being released.

d) The velocity (both magnitude and direction) of the block the first time it is 6.00 cm to
the left of equilibrium.

e) The maximum speed of the block.

Explanation / Answer

a)
F = k × x
and
F = m × g
so
2.5 × 9.8 = k × 0.25
k =98N/m

b)
T = 2 (m/k)
T = 2 ( 2.5 / 98 ) = 0.025 s

For c) , d) and e) , I will first derive a few formula's.

A = amplitude = 0.2 m
= angular velocity = (k/m) = ( 98 / 2.5 ) = 6.2 rad/s

Position at time t :
x = A cos(t)
x = 0.2 cos(6.2t)

Velocity at time t:
v = -A sin(t)
v = -1.4 sin(6.2t)

Acceleration at time t:
a = -A ² cos(t)
a = -9.8 cos(6.2t)

c)
a = -9.8 cos(6.2×0.5)
a = 4.8 m/s² (positive, so to the right)

d)
v = -1.4 sin(6.2×0.6)
v = 0.0.090 m/s (positive, so to the right)

e)
Vmax = A = 9.8*6.2 =60.75 m/s

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