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Oscillatory motion is a special type of periodic motion. In a periodic motion the motion of the body repeats itself in an equal interval of time.
Some of the examples of periodic motion are:
- Motion of the planets around the sun
- Motion of the moon around the earth
- Motion of the mass attached to a spring
- Motion of the bob of simple pendulum
In all these examples, the motion of the body is repeating over a fixed period of time.
Now if we look at the motion of the mass attached to a spring and motion of the bob of a simple pendulum.
- They all move back and forth about a fixed point after a regular interval of time.
- This type of periodic motion is called Oscillatory motion.
| Table of Content |
Key Terms: Oscillatory motion, periodic motion, simple pendulum, simple harmonic motion, velocity, acceleration, frequency, time period.
Read More: NCERT Solutions for Class 11 Physics Motion in a Plane
Oscillatory Motion
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The motion of the body is said to be oscillatory if it moves to and fro about a fixed point after regular intervals of time.
- Oscillatory motion is also called Vibratory motion.
- In oscillatory motion, the body moves to and fro about a fixed point which is known as the mean position or equilibrium position.
- The maximum displacement of the body from its mean position is called amplitude.
- Every oscillatory motion is a periodic motion, but every periodic motion is not oscillatory.
- Motion of the planet around the sun is periodic but not oscillatory.
- To and fro motion of pendulum is oscillatory as well as periodic.
Read More:
| Relevant Concepts | ||
|---|---|---|
| Motion in a Straight Line | Projectile Motion Formula | Motion in a Plane |
| Horizontal Motion | Simple Harmonic Motion | Uniform and Non-Uniform Motion |
Oscillatory Motion Formula
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The simplest form of oscillatory motion is known as Simple harmonic motion (SHM).
A particle is said to execute simple harmonic motion, if it moves to and fro about a mean position under the action of restoring force (F), which is directly proportional to its displacement (x) from the mean position and is always directed towards the mean position.
- In SHM, restoring force (F) ∝ displacement (x)
∴ F = – kx
Where, k is known as force constant. Negative sign shows the direction of restoring force is opposite to that of displacement.
- The time taken by an oscillating body to complete one oscillation is called its time period (T).
- The number of oscillations completed by an oscillating body in one second is called frequency of oscillation (ɳ). The SI unit of frequency is hertz (Hz).
- Relationship between frequency and time period of oscillation is given by
ɳ = \(\frac{1}{T}\)
Mathematically, a simple harmonic motion is expressed as
x = A sin ωt or x = A cos ωt
Where, x = displacement of the body from mean position at any instant t.
A = maximum displacement or amplitude of displacement of the body.
ω = angular frequency = 2πƞ = 2π/T
Velocity of particle in SHM
Velocity of a particle executing simple harmonic motion, is given by
v = ω \(\sqrt{A^2-x^2}\)
- When x = 0, then v = ωA, i.e. velocity will be maximum at x = 0, known as maximum velocity or velocity amplitude
- When x = A, then v = 0, i.e. velocity will be minimum at maximum displacement.
Acceleration of particle in SHM
Acceleration of a particle executing simple harmonic motion, is given by
a = – ω2 x
- When x = 0, then a = 0, i.e. acceleration will be minimum at x = 0.
- When x = A, then a = – ω2A, i.e. acceleration will be maximum at maximum displacement.
Time period and Frequency in SHM
Time period of a particle executing simple harmonic motion is given by
T = 2π \(\sqrt{\frac{x}{a}}\) = 2π \(\sqrt{\frac{displacement}{acceleration}}\)
Time period of simple pendulum is given by
T = 2π \(\sqrt{\frac{l}{g}}\) = 2π \(\sqrt{\frac{length of string}{acceleration due to gravity}}\)
Frequency of a particle executing SHM is given by
ƞ = \(\frac{1}{2 \pi} \sqrt{\frac{a}{x}}\) = 2π \(\sqrt{\frac{acceleration}{displacement}}\)
Oscillatory Motion Examples
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Some examples of oscillatory motion are:
- Motion of the piston of an automobile engine
- Motion of the bob of a simple pendulum
- Motion of the mass attached to a spring
- Vibrations of electrons in the antenna’s of the radio and TV transmitters
- Vibrating strings of musical instruments
- The vibration of tuning forks is also an example of oscillating motion
Check More:
| Related Topics | ||
|---|---|---|
| Uniform Circular Motion | Rotational Kinetic Energy | Angular Momentum |
| Rolling Motion | Moment of Force | Angular Displacement |
Solved Examples
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Ques. On an average human heart beat is found to be 75 times in a minute. Calculate its beat frequency and period.
Ans. Given, number of oscillations = 75 per minute = 75 times in 60 seconds
Frequency, ƞ = \(\frac{number of oscillations}{time}\) = \(\frac{75}{60}\) = 1.25 Hz
Time period, T = \(\frac{1}{\eta}\) = \(\frac{1}{1.25}\) = 0.8 s
That means, in one second it completes 1.25 oscillations, and in 0.8 second it completes 1 oscillation.
Ques. Two simple pendulums of length 1 m and 1.21 m are at their mean position with velocities in the same direction at some instant. After how many oscillations of bigger pendulum they will again be in the same phase?
Ans. Smaller pendulum has length, L1 = 1 m
Therefore, time period, T1 = 2π \(\sqrt{\frac{1}{g}}\)
Bigger pendulum has length, L2 = 1.21 m
Therefore, time period, T2 = 2π \(\sqrt{\frac{1.21}{g}}\)
⇒ \(\frac{T_1}{T_2}\) = \(\frac{2 \pi \sqrt{\frac{1}{g}}}{2\pi \sqrt{\frac{1.21}{g}}}\) = \(\sqrt{\frac{1}{1.21}} = \sqrt{\frac{100}{121}} = \frac{10}{11}\)
⇒ 11 T1 = 10 T2
After completing 10 oscillations of the bigger pendulum, the smaller pendulum will complete 11 oscillations.
Hence, after 10 oscillations of the bigger pendulum, they will again be in the same phase.
Things to Remember
- A body is said to be in oscillatory motion if it moves to and fro about a fixed point after regular intervals of time.
- Every oscillatory motion is a periodic motion, but every periodic motion is not oscillatory.
- An example of oscillatory motion is motion of the bob of a simple pendulum.
- Oscillatory motion is also called Vibratory motion.
- Time taken to complete one oscillation is called the Time period of the oscillation.
- The number of oscillations completed in one second is called frequency of the oscillation.
- Relationship between frequency and time period of oscillation is given by ƞ = \(\frac{1}{T}\)
- The simplest form of oscillatory motion is known as Simple harmonic motion
Sample Questions
Ques. The motion of the earth about its axis is periodic and oscillatory (2 marks)
(A) True
(B) False
Ans. Correct option is (B)
The earth completes its one rotation about its axis every 24 hours. So its motion is periodic. But, it does not move to and from about a fixed point. So its motion is not oscillatory.
Ques. A simple pendulum is attached to the roof of a lift. If the time period of oscillation, when the lift is stationary is T, then the frequency of oscillation when the lift falls freely, will be ___________. (2 marks)
(A) Zero
(B) T
(C) 1/T
(D) Infinity
Ans. Correct option is (A)
Time period of a simple pendulum is given by
T = 2π \(\sqrt{\frac{l}{g}}\) = 2π \(\sqrt{\frac{length of string}{acceleration due to gravity}}\)
Then, frequency of oscillation,
ƞ = \(\frac{1}{T}\) = \(\frac{1}{2 \pi} \sqrt{\frac{g}{l}}\)
When the lift falls freely, then g = 0
⇒ ƞ = \(\frac{1}{2 \pi} \sqrt{\frac{0}{l}}\) = 0
Hence, the frequency of oscillation when the lift falls freely, will be zero.
Ques. There is a simple pendulum hanging from the ceiling of a lift. When the lift is standstill, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then what is the new time period of the pendulum? (3 marks)
Ans. Time period of a simple pendulum is given by
T = 2π \(\sqrt{\frac{l}{g}}\)
⇒ T ∝ \(\sqrt{\frac{1}{g}}\) ⇒ \(\frac{T_1}{T_2}\) = \(\frac{\sqrt{\frac{1}{g_1}}}{\sqrt{\frac{1}{g_2}}} = \sqrt{\frac{g_2}{g_1}}\)
When time period, T1 = T, then resultant acceleration, g1 = g.
Therefore, time period T2, when resultant acceleration, g2 = g/4, is given by
⇒ \(\frac{T}{T_2} \) = \(\sqrt{\frac{g/4}{g}}\) = \(\sqrt{\frac{1}{4}}\) = \(\frac{1}{2}\)
⇒ T2 = 2T
Ques. What is the difference between periodic motion and oscillatory motion? (3 marks)
Ans. In a periodic motion the motion of the body repeats itself in an equal interval of time. While, in an oscillatory motion the body moves to and fro about a fixed point after regular intervals of time.
All oscillatory motion is a periodic motion, but every periodic motion is not oscillatory.
Ques. What is the condition for oscillatory motion? (2 marks)
Ans. The necessary condition for a motion to be an oscillatory motion is that the object executing oscillatory motion should move to and fro about a fixed point in a regular interval of time. This fixed point is known as equilibrium position or mean position.
Ques. What is the time period of a pendulum hanging in a satellite? (2 marks)
Ans. Time period of a simple pendulum is given by
T = 2π \(\sqrt{\frac{l}{g}}\)
In a satellite, g = 0
T = 2π \(\sqrt{\frac{l}{0}}\) = ∞
Hence, the time period of a pendulum hanging in a satellite is infinite.
Ques. There is a simple pendulum hanging from the ceiling of a lift. When the lift is standstill, the time period of the pendulum is T. If the resultant acceleration becomes g/16, then what is the new time period of the pendulum? (3 marks)
Ans. Time period of a simple pendulum is given by
T = 2π \(\sqrt{\frac{l}{g}}\)
⇒ T ∝ \(\sqrt{\frac{1}{g}}\) ⇒ \(\frac{T_1}{T_2}\) = \(\frac{\sqrt{\frac{1}{g_1}}}{\sqrt{\frac{1}{g_2}}} = \sqrt{\frac{g_2}{g_1}}\)
When time period, T1 = T, then resultant acceleration, g1 = g.
Therefore, time period T2, when resultant acceleration, g2 = g/4, is given by
\(\frac{T}{T_2} = \sqrt{\frac{g/16}{g}}=\sqrt{\frac{1}{16}} = \frac{1}{4}\)
⇒ T2 = 4T
Ques. The length of a simple pendulum executing simple harmonic motion is increased by 21%. What is the percentage increase in the time period of the pendulum of increased length? (3 marks)
Ans. Time period of a simple pendulum is given by
T = 2π \(\sqrt{\frac{l}{g}}\)
⇒ T ∝ \(\sqrt{l}\) ⇒ T ∝ l1/2
Therefore, the percentage increase in the time period is given by
⇒ \(\frac{\Delta T}{T} \times 100 = \frac{1}{2} \times \frac{\Delta l}{l} \times 100\)
Given, percentage increase in length, \( \frac{\Delta l}{l} \times 100\) = 21%
⇒ \(\frac{\Delta T}{T} \times 100\) = \(\frac{1}{2}\) x 21% = 10.5%
Ques. Give 5 examples of oscillatory motion. (3 marks)
Ans. The 5 examples of oscillatory motion are:
- The motion of the bob of simple pendulum
- Motion of a child on a swing.
- Motion of piston in an automobiles engines
- Motion of the planets around the sun
- Motion of the hands of the clock
Ques. There is a simple pendulum hanging from the ceiling of a lift. When the lift is standstill, the time period of the pendulum is T. If the resultant acceleration becomes 4g, then what is the new time period of the pendulum? (3 marks)
Ans. Time period of a simple pendulum is given by
T = 2π \(\sqrt{\frac{l}{g}}\)
⇒ T ∝ \(\sqrt{\frac{1}{g}}\) ⇒ \(\frac{T_1}{T_2}\) = \(\frac{\sqrt{\frac{1}{g_1}}}{\sqrt{\frac{1}{g_2}}} = \sqrt{\frac{g_2}{g_1}}\)
When time period, T1 = T, then resultant acceleration, g1 = g.
Therefore, time period T2, when resultant acceleration, g2 = 4g, is given by
⇒ \(\frac{T}{T_2} = \sqrt{\frac{4g}{g}}=\sqrt{4} = 2\)
⇒ T2 = \(\frac{T}{2}\)
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