Simple Harmonic Motion definition, Aptitude formulas with solved examples:
Vibration:
A motion
which repeats itself after a certain interval of time may be called a
vibration. Vibration occurs when a system is displaced from a position of
stable equilibrium.
Simple Harmonic Motion (S.H.M.)
The motion
which repeats itself after an equal interval of time is called periodic motion.
The equal interval is called time period.
It is
expressed in terms of circular sine or cosine functions. The simplest form of
harmonic motion is called Simple Harmonic
Motion. Reciprocating motion is an example of simple harmonic motion.
Simple
harmonic motion can be expressed by the equation;
x =
A sin ωt or, x = A cos ω ………………….eq.(i)
Consider, x = A sin ωt
Where, x =
is the displacement
A = is the amplitude
ω = is the circular frequency
The motion
will be repeated after every 2π/ω time, i.e. time period is equal to 2π/ω
seconds.
The velocity
and acceleration of harmonic motion are obtained by differentiating eq.(i) with
respect to time.
∴ velocity, ẋ = Aω cos ωt = Aω sin (ωt + π/2)
And, acceleration, ẍ = -Aω² sin ωt = Aω² cos (ωt + π/2)
= Aω² sin (ωt + π)
Thus, the
acceleration in a simple harmonic motion is always proportional to its displacement
from the mean position and directed towards the mean position.
Example: 01
A S.H.M.
has amplitude 3 cm and a period of 2 seconds. Determine the maximum velocity
and acceleration.
Solution:
We have, x = A sin ωt, ẋ = Aω sin (ωt + π/2), ẍ = Aω² sin (ωt + π)
Given, A = 3
cm and,
T = 2 sec = 2π/ω
∴ω = 2π/T
= 2π/2 = π radian/sec.
∴Maximum
velocity, ẋmax = Aω = 3 × π = 9.425 cm/sec.
And, Maximum
acceleration = ẍmax = Aω²
= 3 × π² = 29.609 cm/sec².
Example: 02
A harmonic motion has a frequency of 10 hertz and its maximum velocity
is 2.5 m/sec. Determine its amplitude, period and maximum acceleration.
Solution:
Frequency, f = ω/2π Hz.
∴ 10 = ω/2π
∴ ω = 62.832 rad/sec
Time period,
T = 1/f = 1/10 = 0.1 sec.
∴Maximum
velocity, ẋmax = Aω
=> 2.5 =
62.832 × A
=> A = 0.03979
m
= 3.979 cm (Amplitude).
And, Maximum
acceleration = ẍmax = Aω²
= 0.03979 × 62.832² = 157. 08 m/sec²