SHMHard
Question
A particle undergoes simple harmonic motion having time period T. The time taken in 
oscillation is
oscillation is Options
A.3/8T
B.5/8T
C.5/12T
D.7/12T
Solution
3/8 oscillations can split as = 
In one fourth oscillation particle will approach at amplitude position so time taken in one fourth oscillation will be T/4 and time taken in remaining 1/8th oscillation can determine by equation
x = A cos ωt
but x =
so 
= A cosωt
⇒ cos
= cosωt so ωt = 
⇒
t = 
⇒ t = 
then total time taken will be t′ =
In one fourth oscillation particle will approach at amplitude position so time taken in one fourth oscillation will be T/4 and time taken in remaining 1/8th oscillation can determine by equation
x = A cos ωt
but x =
so = A cosωt⇒ cos
= cosωt so ωt = ⇒
t = ⇒ t = then total time taken will be t′ =
Create a free account to view solution
View Solution FreeMore SHM Questions
In the above problem, the frequency of oscillations of the cage will be -...An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected t...Displacement of a particle is x = 3 sin2t + 4cos 2t , the amplitude and the max. velocity will be :...The total energy of a particle executing SHM is directly proportional to the square of the following quantity :...When two mutually perpendicular simple harmonic motions of same frequency, amplitude and phase are superimposed...