Chapter 23 : Simple Harmonic Motion 477
and maximum velocity, vmax = rω = 0.1× 4π = 1.257 m/s Ans.
Maximum acceleration
We also know that maximum acceleration,
amax = ω^2 r = (4π)^2 × 0.1 = 15.79 m/s^2 Ans.
Example 23.7. A particle, moving with simple harmonic motion, performs 10 complete
oscillation per minute and its speed, is 60% of the maximum speed when it is at a distance of 8 cm
from the centre of oscillation,. Find amplitude, maximum acceleration of the particle. Also find
speed of the particle, when it is 6 cm far from the centre of oscillation.
Solution. Given : No. of oscillation/min = 10 and when displacement ( y) = 8 cm velocity
(v) = 60% vmax = 0.6 vmax.
Amplitude of the particle
We know that no. of oscillations per sec
10 1
60 6
==∴ Time-period of the motion(T )6
6s
1==and angular velocity, (^22) rad/s
T 63
πππ
ω===
∴ Linear velocity, (^) vry=ω^22 –
or (^) 0.6vrmax=ω^22 – (8) ...(Q y = 8 cm)
(^) 0.6ω=ωrr^2 – 64 ...(vmax = ωr)
(^) 0.6rr=^2 – 64
Squaring both sides,
0.36r^2 = r^2 – 64 or 0.64r^2 = 64
∴ (^264100)
0.64
r ==
or (^) r== 100 10 cm Ans.
Maximum acceleration of the particle
We know that maximum acceleration of the particle,
2
10 10.97 cm /s^2
max 3
ar
⎛⎞π
=ω =⎜⎟× =
⎝⎠
Ans.
Speed of the particle when it is 6 cm from the centre of oscillation
We know that speed of the particle when it is 6 cm from the centre of oscillation,
(^22) –(10)–(6)8.38cm/s2 2
3
vry
π
=ω = = Ans.