a particle executes s.h.m. of amplitude 4 cm ant T = 4 SECONDS. what is the time teaken by it to move from positive extreme position to half the amplitude ?
Let equation be x=a sin kt (k = angular velocity) when it is at extreme +ve position, x = 0 so, sinkt= 1 , kt = pie/2 t = pie/2kt Since,2pie/k = T = time period, t = T/4
Now put x= a/2, you will get t =pie/6 or 5pie/6 . Take any value. So, taking t=5pie/6, t = 5T/12
time taken = 5T/12 - T/4 = T/6 = 4/6 = 2/3 seconds
Moderation Team
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