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__Simple Harmonic Motion definition, Aptitude formulas with solved examples__:

__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)*
=

*A**ω² 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 = 2

*π/ω*
∴

*ω*= 2*π/T =*2*π/2 = π*radian/sec.
∴Maximum
velocity,

*ẋ*3 ×_{max}= Aω =*π*= 9.425 cm/sec.
And, Maximum
acceleration =

*ẍ*_{max}=*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/*2*π*Hz.
∴ 10 =

*ω/*2*π*
∴

*ω*= 62.832 rad/sec
Time period,

*T = 1/f =*1/10 = 0.1 sec.
∴Maximum
velocity,

*ẋ*_{max}= Aω
=> 2.5 =
62.832 × A

=> A = 0.03979
m

= 3.979 cm (Amplitude).

And, Maximum
acceleration =

*ẍ*_{max}=*Aω² =*0.03979 × 62.832² = 157. 08 m/sec²
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