Dynamics of Harmonic Oscillations

Harmonic oscillations

The oscillations of a physical system results from two basic properties namely elasticity and inertia. Let us consider a body displaced from a mean position. The restoring force brings the body to the mean position.

(i) At extreme position when the displacement is maximum, velocity is zero. The acceleration becomes maximum and directed towards the mean position.

(ii) Under the influence of restoring force, the body comes back to the mean position and overshoots because of negative velocity gained at the mean position.

(iii) When the displacement is negative maximum, the velocity becomes zero and the acceleration is maximum in the positive direction. Hence the body moves towards the mean position. Again when the displacement is zero in the mean position velocity becomes positive.

(iv) Due to inertia the body overshoots the mean position once again. This process repeats itself periodically. Hence the system oscillates.

The restoring force is directly proportional to the displacement and directed towards the mean position.

That is,

F ∝ y

F = -ky             …... (1)

where k is is the force constant. It is the force required to give unit displacement. It is expressed in N m⁻¹.

From Newton’s second law,   F  = ma             …... (2)

−k y = ma

Or, a = (-k/m) y               …... (3)

From definition of SHM acceleration a = −ω²y

The acceleration is directly proportional to the negative of the displacement.

Comparing the above equations we get,

ω = √k/m            …... (4)

Therefore the period of SHM is

T = 2π/ω = 2π√m/k

So,

T =  2π√inertial factor/spring factor            …... (5)

Angular harmonic oscillator

Simple harmonic motion can also be angular. In this case, the restoring torque required for producing SHM is directly proportional to the angular displacement and is directed towards the mean position.

Consider a wire suspended vertically from a rigid support. Let some weight be suspended from the lower end of the wire. When the wire is twisted through an angle θ from the mean position, a restoring torque acts on it tending to return it to the mean position. Here restoring torque is proportional to angular displacement θ.

Hence τ = − C θ                  …... (1)

where C is called torque constant.

It is equal to the moment of the couple required to produce unit angular displacement. Its unit is N m rad⁻¹.

The negative sign shows that torque is acting in the opposite direction to the angular displacement. This is the case of angular simple harmonic motion.

Examples : Torsional pendulum, balance wheel of a watch.

But, τ = Iα                  …... (2)

where τ is torque, I is the moment of inertia and α is angular acceleration.

Thus, angular acceleration,

α = r/I = – Cθ/I        …... (3)

This is similar to α = – ω²y

Replacing y by θ, and a by α we get,

α = – ω²θ = – (C/I) θ

So, ω = √C/I

Thus, period of SHM, T = 2π√I/C

Therefore, frequency, n = 1/T = 1/(2π√I/C) = (1/2π)√(C/I)

Thus,

Refer this video to know more about on, “Dynamic of Harmonic Oscillations”.

Problem (JEE Main):

Let T₁ and T₂ be the time periods of two springs A and B when a mass m is suspended from them separately. Now both the springs are connected in parallel and same mass m is suspended with them. Now let T be the time period in this position. Then

(a) T = T₁ + T₂            (b) T = T₁T₂/ T₁ + T₂

(c) T² = T₁² + T₂²        (d) 1/T² = 1/T₁² + 1/T₂²

Solution:

We know that,

T₁ = 2π√m/k₁

Or, k₁ = 4π²m/T₁²

T₂ = 2π√m/k₂ 

Or, k₂ = 4π²m/T₂²

In parallel, k = k₁ + k₂

Substituting the values of k, k₁ and k₂ we get,

1/T² = 1/T₁² + 1/T₂²

Thus from the above observation, we conclude that, option (d) is correct.

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

(a) 1/ 2π√3         (b) 2π√3 

(c) 2π/√3           (d) √3/2π

Question 2

If the initial displacement is doubled what happens to the maximum kinetic energy of the thingy?

(a) It is unchanged.                            (b) It is doubled.

(c) It is increased by a factor of 4.        (d) We can't tell from the information provided.

Question 3

An object swinging on the end of a string forms a simple pendulum. Some students (and some texts) often cite the simple pendulum's motion as an example of SHM. That is not quite accurate because the motion is really

(a) approximately SHM only for small amplitudes.

(b) exactly SHM only for amplitudes that are smaller than a certain value.

(c) approximately SHM for all amplitudes.

Question 4

The amplitude of the oscillation is determined by

(a) the amount of the initial displacement.

(b) the mass of the thingy and the properties of the spring

(c) the local gravitational field, g.

(d) all of the above.

Question 5

Which one of the following statements is not true for a body vibrating in simple harmonic motion when damping is present?

(a) The damping force is always in the opposite direction to the velocity.

(b) The damping force is always in the opposite direction to the acceleration.

(c) The presence of damping gradually reduces the maximum potential energy of the system

(d) The presence of damping gradually reduces the maximum kinetic energy of the system.

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