A block of mass m=0.750 kg is fastened to an unstrained horizontal spring whose

Question

A block of mass m=0.750 kg is fastened to an unstrained horizontal spring whose spring constant is k=82 N/m. The block is given a displacement of +0.120 m, where the + sign indicates that the displacement is along the +x-axis, and then released from rest.

(a) What is the force (magnitude and direction) that the spring exerts on the block just before the block is released?

(b) find the frequency ((number of cycles)/time) of the resulting oscillatory motion.

(c) Maximum speed of the block?

(d) Magnitude of the maximum acceleration of the block?

Explanation / Answer

(a) We know that the force exerted by the spriing = kx
where k is spring constant and x is the displacement of the spring.
Hence Force = 82*0.12 = 9.84 N
(b) We know that the frequency in the spring mass system is given by
f= (1/2Pi)(K/m)1/2
f = (1/2Pi)(82/0.75)1/2 = 1.664 Hz
(c) Maximum speed of the block (Vmax)
We know that the energy at the mean position is = (1/2)mVmax2
Which is equal to the potential energy of the spring at extreme position = (1/2)kx2
(1/2)mVmax2 = (1/2)kx2
Vmax = 1.2547 m/s
(d) maximum acceleration is given by = -w2A
where w = (K/m)1/2
and A = 0.12 m
acceleration = -(K/m)*0.12 = - 13.12 m/s2
-ve sign indicate that the acceleration is toward mean position.

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