**Beats**

Beats is the phenomenon caused by two sound waves of nearly same frequencies and amplitudes travelling in the same direction.

For example:-

- Tuning of musical instruments like piano, harmonium etc. Before we start playing on these musical instruments they are set against the standard frequency. If it is not set a striking noise will keep on coming till it is set.

Mathematically

- Consider only the time dependent and not the position dependent part of the wave.
- s
_{1}=a cos ω_{1}t and s_{2}=a cos ω_{2}t; where amplitude and phase of the waves are same,but the frequencies are varying. Also considering ω_{1}> ω_{2}. - When these 2 waves superimpose s= s
_{1}+ s_{2}=a[cos ω_{1}t + cos ω_{2}t] - By simplifying , 2a (cos(ω
_{1}– ω_{2})/2)t cos(ω_{1}+ ω_{2})/2)t) - =>ω
_{1}– ω_{2}is very small as ω_{1}> ω_{2}.Let (ω_{1}– ω_{2 })=ω_{b} - =>ω
_{1}+ ω_{2 }is very large. Let (ω_{1}+ ω_{2 })=ω_{a} **s= 2a cos ω**_{b}t cos ω_{a}t- cosω
_{a}t will vary rapidly with time and 2acosω_{b}t will change slowly with time. - Therefore we can say 2acosω
_{b}t = constant. As a result 2acosω_{b}t = amplitude as it has small angular variation.

** Problem:- **Two sitar strings A and Bplaying the note ‘Dha’ are slightly out oftune and produce beats of frequency 5 Hz.The tension of the string B is slightlyincreased and the beat frequency is foundto decrease to 3 Hz. What is the originalfrequency of B if the frequency of A is 427 Hz?

** Answer**:- Increase in the tension of a stringincreases its frequency. If the original frequencyof B (ν

_{B}) were greater than that of A (ν

_{A}), furtherincrease in ν

_{B}should have resulted in anincrease in the beat frequency. But the beatfrequency is found to decrease. This shows thatν

_{B}< ν

_{A}. Since ν

_{A}– ν

_{B}= 5 Hz, and ν

_{A}= 427 Hz, weget ν

_{B}= 422 Hz.

**Doppler’s Effect**

- Doppler Effect is the phenomenon of motion-related frequency change.
- Consider if a truck is coming from very far off location as it approaches near our house, the sound increases and when it passes our house the sound will be maximum. And when it goes away from our house sound decreases.
- This effect is known as Doppler Effect.
- A person who is observing is known as
**Observer**and object from where the sound wave is getting generated it is known as**Source**. - When the observer and source come nearer to each other as a result waves get compressed. Therefore wavelength decreases and frequency increases.
**Case 1**:- stationary observer and moving source- Let the source is located at a distance L from the observer.
- At any time t
_{1}, the source is at position P_{1}. - Time taken by the wave to reach observer =L/v where v=speed of the sound wave.
- After some time source moves to position P
_{0}in time T_{0}. - Distance between P
_{1}and P_{0}=v_{s}T_{o}where v_{s}is the velocity of the source. - Let t
_{2}be the time taken by the second wave to reach the observer- Total time taken by the for the second wave to be sent to the observer = T
_{o}+( L+v_{s}T_{o})/v - Total time taken by the for the third wave to be sent to the observer=2T
_{o}+( L+2v_{s}T_{o})/v - Therefore for nth point t
_{n+1}=nT_{o}+( L+nv_{s}T_{o})/v - =>In time t
_{n+1}the observer captures n waves.

- Total time taken by the for the second wave to be sent to the observer = T
- Total time taken by the waves to travel Time period T= (t
_{n+1}– t_{1})/n - =T
_{o}+(v_{s}T_{o})/v =>T=T_{o}(1+v_{s}/v) - Or v= 1/T
- =>v = v
_{0}(1+v_{s}/v)^{-1} - By using binomial Theorem,
**v= v**_{0 }(1- v_{s}/v) - If the source is moving towards the observer the expression will become
**v= v**_{0 }(1+ v_{s}/v) **Case 2**:- moving observer and stationary source- As the source is not moving therefore v
_{s}is replaced by -v_{0}. - Therefore
**v= v**_{0 }(1+ v_{0}/v)