A force of magnitude 40.0 N stretches a vertical spring a distance 0.240 m . Par

Question

A force of magnitude 40.0 N stretches a vertical spring a distance 0.240 m .

Part A

What mass must be suspended from the spring so that the system will oscillate with a period of 1.10 s ?

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Part B

If the amplitude of the motion is 5.50×102 m and the period is that specified in part (a), where is the object at a time 0.370 s after it has passed the equilibrium position, moving downward? (Take the upward direction positive.)

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Part C

What force (magnitude) does the spring exert on the object when it is a distance 3.30×102 m below the equilibrium position, moving upward?

Take the free fall acceleration to be 9.80 m/s2 .

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Part D

Find the direction of this force.

Find the direction of this force.

Explanation / Answer

This is just using the expression for the period T:

T = 2 (m/k)

where m is the mass and k is the spring constant.

k = (40 N) / (0.240 m) = 166.67 N/m

Plugging in the given values:

1.1 = 2 (m/166.67)

(1.1 / 2)² = m/166.67

m = 5.11 kg

B.The 'shape' of the sin() and cos() functions are the same, so you can use either if you stick in an appropriate phase shift , but think about the starting conditions for most mechanical oscillators, when t = 0.

To describe the displacement (x), the general expression would be either:

x = Asin(2pi t )

{ is the angular frequency = 2f = 2/T }

x=5.50×102sin(2pix.370)=2.23x10^-3m (moving upward since theta>pi/2)

3. Force does the spring streched 3.30×102 m from equillibrium

Fs=KdetaX

K=40/.24=166.67

=166.67x3.30×102=5.5N

D.

the direction of this force.

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