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Oscillations - Periodic and Oscillatory Motions

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Periodic Motion: Any motion that repeats itself at regular intervals of time is called periodic motion. The fixed interval of time is called the period TT. Examples include the rotation of the Earth about its axis and the motion of hands of a clock.

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Oscillatory Motion: A type of periodic motion in which an object moves to-and-fro or back-and-forth about a fixed mean position. Every oscillatory motion is periodic, but every periodic motion (like circular motion) is not necessarily oscillatory.

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Period (TT): The smallest interval of time after which the motion repeats. Its SI unit is the second (ss).

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Frequency (ν\nu or ff): The number of repetitions or cycles per unit time. It is the reciprocal of the period: ν=1T\nu = \frac{1}{T}. The SI unit is Hertz (HzHz or s−1s^{-1}).

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Angular Frequency (ω\omega): Represents the rate of change of phase angle. It is related to frequency by ω=2πν\omega = 2\pi\nu. The SI unit is rad/srad/s.

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Harmonic Motion: A periodic motion that can be represented by a single sine or cosine function, such as y=Asin⁡(ωt)y = A \sin(\omega t) or y=Acos⁡(ωt)y = A \cos(\omega t).

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Phase and Phase Constant: The parameter (ωt+ϕ)(\omega t + \phi) is the phase of the motion, which describes the state of motion at any time tt. The term ϕ\phi is the phase constant or initial phase at t=0t = 0.

📐Formulae

T=1νT = \frac{1}{\nu}

ω=2πν=2πT\omega = 2\pi\nu = \frac{2\pi}{T}

x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi)

v(t)=dxdt=−ωAsin⁡(ωt+ϕ)v(t) = \frac{dx}{dt} = -\omega A \sin(\omega t + \phi)

💡Examples

Problem 1:

A simple pendulum completes 4040 full oscillations in 8080 seconds. Calculate its time period (TT) and frequency (ν\nu).

Solution:

The time period TT is the time for one oscillation: T=Total TimeNumber of Oscillations=8040=2 sT = \frac{\text{Total Time}}{\text{Number of Oscillations}} = \frac{80}{40} = 2\text{ s} The frequency ν\nu is the reciprocal of the period: ν=1T=12=0.5 Hz\nu = \frac{1}{T} = \frac{1}{2} = 0.5\text{ Hz}

Explanation:

By definition, the period is the time per cycle, and the frequency is the cycles per unit time.

Problem 2:

Determine the angular frequency ω\omega of a particle that has a frequency of 50 Hz50\text{ Hz}.

Solution:

Given ν=50 Hz\nu = 50\text{ Hz}. The formula for angular frequency is: ω=2πν\omega = 2\pi\nu ω=2×3.14159×50\omega = 2 \times 3.14159 \times 50 ω=314.16 rad/s\omega = 314.16\text{ rad/s}

Explanation:

Angular frequency is obtained by multiplying the linear frequency by 2π2\pi to convert cycles to radians.

Problem 3:

An experiment measures the time for 100100 oscillations twice. The first reading is 155.45 s155.45\text{ s} and the second reading is 120.20 s120.20\text{ s}. Calculate the difference in time using vertical arithmetic.

Solution:

To find the difference: 155.45−120.2035.25\begin{array}{r} 155.45 \\ - 120.20 \\ \hline 35.25 \end{array} The difference is 35.25 s35.25\text{ s}.

Explanation:

Standard subtraction is used to find the interval between two measured time values.