An oscillator consists of a block of mass 512 g connected to a spring. When set into oscillation with amplitude 34.7 cm, it is observed to repeat its motion every 0.484 s. Find (a) the period, (b) the frequency, (c) the angular frequency, (d) the force constant, (e) the maximum speed, and (f) the maximum force exerted on the block.

8 years ago
(a) Since the motion is repeated in every 0.484 s, therefore the period of oscillation T will be,T = 0.484 s
(b) To find out the frequency f of oscillation, substitute T = 0.484 s in the equation f = 1/T,f = 1/T
= 1/0.484 s
= (2.07/s) (s / 1/Hz)
= 2.07 Hz
From the above observation we conclude that, frequency of the oscillation would be 2.07 Hz.
(c) To find out the angular frequency w of the oscillation, substitute 2.07 Hz for f in the equation w = 2πf,
w = 2πf
= 2×3.14×2.07 Hz
= 2×3.14×(2.07 Hz× (1/1 sec)/1 Hz)
= 2×3.14×2.07/s
From the above observation we conclude that, angular frequency w of the oscillation would be 13.0 rad/s.
(d) To find out the force constant k, substitute 512 g for m and 13.0 rad/s for w in the euation k = mw2

pa1
357 Points
8 years ago
2(a) Since the motion is repeated in every 0.484 s, therefore the period of oscillation T will be,T = 0.484 s(b) To find out the frequency f of oscillation, substitute T = 0.484 s in the equation f = 1/T,f = 1/T= 1/0.484 s= (2.07/s) (s / 1/Hz)= 2.07 HzFrom the above observation we conclude that, frequency of the oscillation would be 2.07 Hz.(c) To find out the angular frequency w of the oscillation, substitute 2.07 Hz for f in the equation w = 2pf,w = 2pf= 2×3.14×2.07 Hz= 2×3.14×(2.07 Hz× (1/1 sec)/1 Hz)= 2×3.14×2.07/s= 13.0 rad/sFrom the above observation we conclude that, angular frequency w of the oscillation would be 13.0 rad/s.(d) To find out the force constant k, substitute 512 g for m and 13.0 rad/s for w in the euation k = mw
pa1
357 Points
8 years ago
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