Important Simple Harmonic Motion Formulas for JEE

Simple harmonic motion is one of the important topics included in the periodic motion chapter. We know that any motion which repeats itself after regular intervals of time is known as a periodic motion. There are several examples of periodic motion, for example, the motion of planets around the Sun, the motion of hands of a clock, motion of the moon around the earth, the motion of Halley’s comet around the Sun observable on the Earth once in 76 years, etc. 

If the restoring force of the restoring torque acting on the dynamic body or the system which is in oscillatory motion will be directly proportional to the displacement of the body and is always directed towards equilibrium or the mean position. Then, such kind of motion is known as Simple Harmonic Motion which is also abbreviated as SHM. Here, we will discuss important simple harmonic motion formulas for jee such as the simple harmonic motion formula or the oscillation formula, and other related topics.. 

Simple Harmonic Motion Formula

• If an object is subjected to a motion such that it moves back and forth periodically or repeatedly about the equilibrium position, then the body is said to be executing oscillatory motion. 

• The motion of bob attached to a pendulum, string vibrations of a guitar, the vibrations obtained with a tuning fork, oscillations of a spring whose one end is attached to a wall, etc… are the few general examples of oscillatory motion.

• In all the above illustrations mentioned the vibrations of bodies we must notice that the path of vibration is always directed towards the mean or equilibrium position i.e., the body is always trying to stay in the equilibrium state. This kind of oscillatory motion is known as the simple harmonic motion (SHM).

• Therefore, the simple harmonic motion can be defined as an oscillatory motion about a fixed point in which the restoring force is always directly proportional to the displacement and directed always towards that fixed point or mean position.

⇒ F = -ky

Where,

k - The proportionality constant and is equal to mω\[^{2}\]

y - The mean displacement

Oscillation Formula

Now, let us have a look at the oscillation formula or important simple harmonic motion formulas for jee as listed below. All these simple harmonic formulas are of importance from the exam point of view. 

1. General Equation of Simple harmonic motion:

Displacement y = A sin(ωt + φ)

Where,

A - The amplitude of the oscillation i.e., maximum vibration from the mean position

ω - The angular frequency

φ - The phase angle

2. Angular Frequency (ω):

The angular frequency of the body executing simple harmonic motion is given by

⇒ ω = 2\[\pi\]f or ⇒ ω = \[\frac{2 \pi}{T}\] 

Where,

f - The frequency

T - The time period

3. Time Period (T):

The time period is inversely proportional to the frequency of the body. The time period of the oscillatory body is the total time taken to complete one cycle and it is given by:

⇒ T = \[\frac{2 \pi}{\omega }\]

Where,

ω - The angular frequency of the oscillating body

3. Hooke’s law:

Hooke’s law states that the total restoring force will be directly proportional to the displacement of the body.

⇒ F = -ky

y - The total displacement for small elongations

k - The spring constant

4. Velocity:

Since we know that the velocity of a body is the rate of change of its displacement, therefore we get:

⇒ v = \[\frac{dy}{dt}\] = \[\frac{d}{dt}\]{A sin(ωt + φ)}

⇒ v =  A {cos (ωt + φ)ω}

⇒ v = ω A cos(ωt + φ)

Where,

A - The amplitude of the oscillation i.e., maximum vibration from the mean position

ω - The angular frequency

φ - The phase angle

5. Acceleration:

Acceleration is the rate of change of velocity, therefore, the acceleration of the oscillating body is given by:

⇒ a = \[\frac{dv}{dt}\] = \[\frac{d}{dt}\] {ω A cos (ωt + φ)}

⇒ a = Aω {- sin (ωt + φ)ω}

⇒ a = - ω\[^{2}\] A sin (ωt + φ) = - ω\[^{2}\]y 

Where,

A - The amplitude of the oscillation i.e., maximum vibration from the mean position

ω - The angular frequency

φ - The phase angle

6. Potential Energy:

The potential energy formula is given by:

⇒ U = \[\frac{1}{2}\]ky\[^{2}\]

Where,

y - The total displacement for small elongations

k - The spring constant

7. Kinetic Energy:

⇒ K.E = \[\frac{1}{2}\] k(A\[^{2}\] - y\[^{2}\])

8. Total Energy:

We know that total energy is given by the sum of potential energy and kinetic energy, therefore the total energy is given by:

E\[_{total}\] = P.E + K.E = \[\frac{1}{2}\]kA\[^{2}\]

Where,

k = mω\[^{2}\] - The spring constant

9. The Time Period of the Simple Pendulum:

⇒ T = 2\[\pi\]\[\sqrt{\frac{l}{g}}\]

These are the shm formula list and for detailed derivations of the oscillation formula or simple harmonic motion formula students are advised to go through the shm notes for jee. The shm IIT JEE notes will boost your exam preparation. Before starting with the examples and important simple harmonic motion formulas for jee students should thoroughly go through the waves and oscillations IIT JEEchapter.

Examples:

1) A Particle of Mass 0.2 Kg Executes Simple Harmonic Motion Under a Force F = -10 X N. If the Speed of Particle About the Mean Position is 8m/s. Then Calculate the Amplitude of Oscillation Obtained from the Body.

Sol:

Given,

The mass of the body = m = 0.2 kg

The restoring force acting on the body = F = - 10 x N

The speed of the particle about mean position = v = 8 m/s

We are asked to determine the amplitude of the oscillation. Before we start with our calculation we have to first calculate the angular frequency of the oscillation.

We know that the spring constant k is given by:

⇒ k = mω\[^{2}\]

⇒ ω = \[\sqrt{\frac{k}{m}}\] = \[\sqrt{\frac{10}{0.2}}\]

⇒ ω = 7.017 Hz

Now, according to the conservation of mechanical energy, we know that,

⇒ total energy = kinetic energy

⇒\[\frac{1}{2}\]kA\[^{2}\] = \[\frac{1}{2}\]mv\[^{2}\]

⇒ A = v\[\sqrt{\frac{m}{k}}\] = \[\frac{v}{\omega }\]

⇒ A = \[\frac{8}{7.017}\]

⇒ A = 1.14 m

Therefore, the amplitude of the given oscillation is 1.14m.

2) A Body in Simple Harmonic Motion has an Amplitude of 6cm and a Period of 0.3 S. Calculate the Acceleration and the Velocity of the Body When the Displacement is (a) 6 Cm (b) 0 Cm.

Sol:

Given,

The amplitude of the particle simple harmonic motion = A = 6cm

The total time period of the particle = T = 0.3 sec

The angular frequency of the particle simple harmonic motion = ω = \[\frac{2 \pi}{T}\] = \[\frac{2 \pi}{0.3}\] = 20.94 rad/sec ≈ 21 rad/sec

a) The displacement of the particle executing simple harmonic motion = y = 6cm

We are asked to determine the velocity of the body, let us assume that the particle is at its mean position. Then we know that, the velocity of the particle is given by:

⇒ v = ω\[\sqrt{A^{2} - y^{2}}\]

Where,

ω - The angular frequency of the particle

A - The amplitude of the oscillation

y - The displacement of the particle executing simple harmonic motion

Substituting all the known values in the above expression, we get:

⇒ v = ω\[\sqrt{A^{2} - y^{2}}\] = 21 \[\sqrt{6^{2} - 6^{2}}\]

⇒ v = 0 m/s

b) The displacement of the particle executing simple harmonic motion = y = 0 cm

We are asked to determine the velocity of the body, let us assume that the particle is at its mean position. Then we know that the velocity of the particle is given by:

⇒ v = ω\[\sqrt{A^{2} - y^{2}}\]

Where,

ω - The angular frequency of the particle

A - The amplitude of the oscillation

y - The displacement of the particle executing simple harmonic motion

Substituting all the known values in the above expression, we get:

⇒ v = ω\[\sqrt{A^{2} - y^{2}}\] = 21\[\sqrt{6^{2} - 0^{2}}\]

⇒ v = 126 m/s

Therefore, the velocity of the body when the displacement is 6cm and 0cm is 0 m/s and 126 m/s respectively.