Chapter 23 : Simple Harmonic Motion „„„„„ 475

Time required in passing between the two points,
t = t 1 – t 2 = 0.59 – 0.26 = 0.33 s Ans.
Example 23.5. A body performing simple harmonic motion has a velocity of 12 m/s when
the displacement is 50 mm, and 3 m/s when the displacement is 100 mm, the displacement being
measuerd from the mid-point. Calculate the frequency and amplitude of the motion. What is the
acceleration when the displacement is 75 mm?

Solution. Given : Velocity (v 1 ) = 12 m/s, when displacement ( y 1 ) = 50 mm = 0.05 m ; and
velocity (v 2 ) = 3 m/s, when displacement ( y 2 ) = 100 mm = 0.1 m

Amplitude of the motion

Let r = Amplitude of the motion, and

ω = Angular displacement of the body,

We know that velocity of the body,

vry=ω^22 –

∴ 12 =ω rr^222 – (0.05) =ω – 0.0025 ...(i)

Similarly 3–(0.1) –0.01=ω rr22 2=ω ...(ii)

Dividing equation (i) and (ii)

22

22

12 – 0.0025 – 0.0025

or 4

(^3) – 0.01 – 0.01
rr
rr
ω

ω
2
2

• 0.0025
  16
  –0.01

r

r

= ...(Squaring both sides)

16 r^2 – 0.16 = r^2 – 0.0025

15 r^2 = – 0.0025 + 0.16 = 0.1575

∴

2 0.1575 0.0105

15

r ==

or r = 0.1025 m Ans.

Frequency of the motion

Substituting the value of r in equation (i),

(^12) =ω (0.1025) – 0.0025^2 =ω×0.09
∴^12 133.3 rad/sec
0.09
ω= =
We know that frequency of the motion,
1 133.3
21.2 Hz
2
N
T
== =
π
Ans.
Acceleration when the displacement is 75 mm
We know that acceleration of the body when y is 75 mm or 0.075 m,
a = ω^2 y = (133.3)^2 × 0.075 = 1332.6 m/s^2 Ans.