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CalibratePhysics

5.3

Linear Simple Harmonic Motion (S.H.M.)

Oscillations · Maharashtra HSC

Theory

The spring–block oscillator

Diagram (Fig. 5.1): a rectangular block of mass mm rests on a smooth frictionless horizontal surface. One end of a light spring is attached to a rigid wall on the left and the other end to the block. Three positions are drawn — (a) the block pulled to the right of the equilibrium position with the spring force arrow pointing left, (b) the equilibrium position with no force on the block, and (c) the block pushed to the left with the spring compressed and the force arrow pointing right.

Setting up the motion

  • Pull the block towards the right and release it. It begins its to and fro motion on either side of its equilibrium position. This motion is linear simple harmonic motion.
  • In the equilibrium position (b) the spring exerts no force on the block.
  • Displaced towards the right (a), the spring, by virtue of its elastic properties, tends to regain its original shape and exerts a restoring force on the block directed towards the left. The force is proportional to the displacement but opposite in direction:

f=−kxf=-kx where kk is a constant depending on the elastic properties of the spring, called the force constant, and the negative sign shows force and displacement are oppositely directed.

  • Displaced towards the left (c), the force exerted by the spring is directed towards the right, again with magnitude proportional to the displacement. So f=−kxf=-kx serves as the equation of motion in both directions.

Acceleration

If mm is the mass of the block, a=fm=−(km)xa=\dfrac{f}{m}=-\left(\dfrac{k}{m}\right)x The acceleration is also proportional to the displacement and its direction is opposite to that of the displacement; the force and the acceleration are both directed towards the mean (equilibrium) position.

One complete oscillation, described

  • As the block moves towards the mean position its speed increases because of the acceleration, while its displacement decreases.
  • At the mean position the displacement, and hence the force and acceleration, are zero; the speed is maximum and the kinetic energy attains its maximum value.
  • The block therefore does not stop at the mean position but continues beyond it towards the left. The spring is compressed and exerts a restoring force towards the right; force and displacement are again oppositely directed.
  • This opposing force retards the block, its speed falls and finally becomes zero. Here the displacement from the mean position and the restoring force are maximum.
  • The force now accelerates the block back towards the equilibrium position, and the process repeats, causing the block to oscillate on either side of its mean position.

Definition

Linear S.H.M. is defined as the linear periodic motion of a body, in which force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.

Conditions required for S.H.M.

  • The oscillation of the particle is about a fixed point.
  • The net force or acceleration is always directed towards that fixed point.
  • The particle comes back to the fixed point because of the restoring force.

A complete oscillation is when the object goes from one extreme to the other and back to the initial position.

Harmonic and non-harmonic oscillation

  • Harmonic oscillation: an oscillation that can be expressed in terms of a single harmonic function, such as x=asin⁡ωtx=a\sin\omega t or x=acos⁡ωtx=a\cos\omega t.
  • Non-harmonic oscillation: an oscillation that cannot be expressed in terms of a single harmonic function. It may be a combination of two or more harmonic oscillations, such as x=asin⁡ωt+bsin⁡2ωtx=a\sin\omega t+b\sin2\omega t.

Two demonstrations of linear S.H.M.

  • A hydrometer immersed in a glass jar filled with water floats vertically in its equilibrium position. If it is slightly depressed and released, it bobs up and down performing linear S.H.M.
  • A U-tube is filled with a sufficiently long column of mercury. With both arms exposed to the atmosphere the levels are the same. If the level in one arm is depressed slightly and released, the level in each arm moves up and down about the equilibrium position, performing linear S.H.M.

Standard results

  • f=−kxf=-kx; kk is the force constant, the restoring force per unit displacement, in N/m
  • a=−(km)x=−ω2xa=-\left(\dfrac{k}{m}\right)x=-\omega^2x, with ω2=km\omega^2=\dfrac{k}{m}
  • T=2πmkT=2\pi\sqrt{\dfrac{m}{k}} and n=12πkmn=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}
  • Since the whole motion lies along one straight line, displacement is measured from the mean position and only ++ and −- signs are needed

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

  1. 1

    A body of mass 1 kg is made to oscillate on a spring of force constant 25×10325\times10^3 dyne/cm. Calculate the magnitude of angular velocity and frequency of vibrations of the body.

    Maharashtra HSC board question — asked in July 2019Answer 1 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

  1. Given: m = 1 kg, k=25×103k=25\times10^3 dyne/cm = 25 N/m To find: Angular velocity (ω), Frequency (n) Formulae: i. ω=km\omega=\sqrt{\dfrac{k}{m}} ii. ω=2πn\omega=2\pi n Calculation: From formula (i), ω=251=5\omega=\sqrt{\dfrac{25}{1}}=5 rad/s From formula (ii), n=ω2π=52π=2.53.14=0.7962n=\dfrac{\omega}{2\pi}=\dfrac{5}{2\pi}=\dfrac{2.5}{3.14}=0.7962 Hz Ans: The angular velocity of the body is 5 rad/s and the frequency of vibration is 0.7962 Hz.