Simple Harmonic Motion Aptitude

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.

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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, ẋ_max = Aω = 3 × 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/2π Hz.

∴ 10 = ω/2π

∴ 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