A 40.0-N force stretches a vertical spring 0.250 m. (a) What mass must be suspen

Question

A 40.0-N force stretches a vertical spring 0.250 m. (a) What mass must be suspended form the spring so that the system will oscillate with a period of 1.0 s? (b) If the amplitude of the motion is 0.050 m and the period is that specified in part (a) where is the object and in whatdirection is it moving 0.35 s after it has passed the equilibrium position, moving downward? Take the positive direction to beupward. (c) What force (magnitude and direction) does the spring exert onthe object when it is 0.030 m below theequilibrium position, moving upward? I need help with part B and C, the text book solution didnt help me

Explanation / Answer

A 40.0 N force stretches a spring by 0.250 m.

As F = kx (in terms of magnitude only)

Hence 40.0 = k(0.250) and k = 160 N/m

Then as T = 2 (m/k)1/2

So m = kT2/42 = (160)(1.0)2/4(3.14)2 = 4.05696 kg

m = 4.05696 kg

now the instantaneous displacement of the mass from equilibrium position is given by y = A sin (2t/T) where A =0.050 m and T = 1.0 s

also as time period is 1.0 s hence it must be kept in mind that the mass take 0.25 s to go from equilibrium position to extreme position and 0.25 s to go from extreme position to equilibrium position.

So when the mass passes the equilibrium position, moving downward. It takes 0.25 s to reach extreme down and then return from there. Hence after t = 0.35 s, particle will be on its way to equilibrium position, moving upward and it displacement will be

y = A sin (2t/T) = (0.050)[sin {2(0.35/1.0)}] = 0.04045 m from the equilibrium position downward.

Hence particle displacement at t = 0.35 s, is -0.04045 m and it is moving upwards at that instant.

Also the magnitude of force exerted by the spring on the mass when it is at x = 0.030 m from the equilibrium position is

F = kx = (160)(0.030) = 4.8 N

Clearly the force acts towards the equilibrium position that means it is acting upwards hence F = 4.8 N.

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