Question

A particle performs S.H.M. of period 12 second and amplitude 8cm. If initially the particle is at the positive extremity, how much time will it take to cover a distance of 6 cm from the extreme position ? (Ans: 2.517s)

Answer 1

given, period , T = 12 sec
Amplitude , A = 8cm
so, angular frequency,$\omega$ = 2π/T = 2π/12 = π/6 rad/s

When the particle covers a distance of 6cm from the positive extremity. Its displacement from the mean position is x = 8-6 = 2cm

We know, equation of SHM is given by
$x=Acos\omega t$[from extreme position, ]

x =8sin(π/6)t

at x = 2cm

2 = 8cos(π/6)t

1/4 = cos(π/6)t

75°52' × π/180° = (π/6) × t

t = 2.517 sec

Answer 2

Answer:

t=2.517s

Explanation:

refer attachment for solution