Waves - Class 11 Medical Physics - Extra Questions

Given,

N lags M by 0.1 s which is equivalent to 

$$\phi =2\pi (\dfrac{0.1}{0.4})$$
Here N is leading M by $$\frac{1}{4^{th}}$$ of a phase i.e.
$$\dfrac{1}{4} \times (2 \pi)=+\dfrac{\pi}{2}$$

$$answer: f_b=345Hz ,f_c=341$$ or $$ 349Hz$$

$$|f_a-f_b|=5$$

$$|f_b-f_c|=4$$     

since on adding wax frequency of A reduced , the difference between frequency of A and B  also reduced this means frequency A >frequency B

now ,

$$f_a-f_b=5$$

$$350-f_b=5$$

$$f_b=345$$

if $$f_b>f_c$$

$$f_b-f_c=4$$     

$$f_c=341Hz$$

Also  $$\nu' \propto  \sqrt{T}$$  where $$T$$ is the tension in the wire

Beat frequency    $$b   =   |\nu - \nu'|$$           ..........(1)

Now it is given that  beat frequency is  $$f$$

Thus       $$f = \nu - \nu'$$        .........(2)

Case A:   Loading wax on tuning fork will decrease   $$\nu$$

Let    $$n$$  be the new frequency of fork after applying wax on it.

Thus it may possible that   $$\nu'< n< \nu$$ which in result decrease $$f$$ and for   $$n<(\nu,   \nu')$$ implies beat frequency may increase  (using 1) depending on their values.

Hence beat frequency may decrease and increase.

Case B:   Filing the prongs of the fork will increase its frequency.

Thus from (2)   beat frequency must increase.

Case C :  Increasing the tension in the wire will increase the frequency of the string waves. Thus it may possible that   $$ \nu' < n < \nu$$ which in result increase $$f$$ and    $$n > (\nu' ,   \nu)$$ implies beat frequency may decrease  (using 1) depending on their values.

Hence beat frequency may decrease and increase.

Case D:       Decreasing the tension in the wire will decrease the frequency of the string waves. Thus from (2)   beat frequency must increase.

given , frequency of fork $$=256Hz$$ ,

           beat frequency $$=4$$ beats/ second ,

therefore , $$n\pm256=4$$ ,

or              $$n=4\pm256=260$$ or $$252Hz$$ ,

   when the string is shortened , its fundamental frequency increases due to law of length and it is given that beat frequency is decreasing on shortening the string's length .

   If fundamental frequency of string is 260Hz , then beat frequency will increase when length is shortened (string frequency is increased) but it is not happening therefore string's frequency is 252Hz .

Now , after shortening the length of string , there are no beats it means , fundamental frequency of string is now 256 Hz ,

we have , initial length $$l=25cm=0$$ ,

length after shortening $$=l'(let)$$ ,

initial frequency $$n=252Hz$$ ,

final frequency   $$=256Hz$$ ,

therefore , by law of  length ,

                 $$n\propto 1/l$$ ,

                 $$l'/l=252/256$$ ,

or              $$l'=l(252/256)=25(252/256)=24.60cm$$ ,

hence , $$l'l'=25-24.60=0.4cm$$

$$answer: 387Hz$$

since on adding wax to unknown tuning fork the frequency $$f_1$$ reduced 

and also beats reduced this means that $$f_1>f_2$$

$$f_1=387Hz$$

$$\Rightarrow v\propto \sqrt T$$

let $$v=\beta \sqrt T$$..........(1)

let $$LB$$ the length of rod at $$0^oC$$

$$L^{\prime}=L(1+\alpha T)\quad \left\{ L^{\prime}=\right. $$ at $$TC$$

$$\begin{cases} L_{ 0 }=L(1+\alpha T_{ 0 }) \\ v_{ 0 }=\beta \sqrt { T_{ 0 } }  \end{cases}$$

now, for resonance

$$\dfrac{k\lambda_0}{4}=L_0$$

$$f_0=\dfrac{k}{4}\dfrac{v_0}{L_0}$$

$$\Rightarrow \boxed{n=\dfrac{k}{4}\dfrac{\beta \sqrt{T_0}}{L(1+\alpha T_0)}}$$.......(iii) 

$$L_1=L_0(1+\alpha dT), \alpha_1=\beta \sqrt{T_0+dt}$$

$$\Rightarrow \left\{ L_1=L(1+\alpha T_0)(1-(\alpha dT)\right\}$$

$$f_1=\dfrac{k}{4}\dfrac{v_1}{L_1}=\dfrac{k}{4}\left(\dfrac{\beta\sqrt{T_0+dt}}{L(1+\alpha T_0)(1+\alpha dt)}\right)$$

$$f_1=\dfrac{k\beta \sqrt{T_0}}{4L(1+\alpha T_0)}\left(\sqrt{1+\dfrac{dt}{T_0}}(1-\alpha dT)\right)$$

using binomial approximation

$$\boxed{f_1=n\left(1+\dfrac{1}{2}\dfrac{dT}{T_0-\alpha dT}\right)}$$......(iii)

$$\Rightarrow \boxed{beat\ frequency \Rightarrow (iii)-(ii)=n\left(\dfrac{1}{2}\dfrac{dT}{T_0}-\alpha dT\right)}$$

Applications

$$\left( a \right) $$ Flow measurement-Instruments such as Lase doppler velocimeter(LDV), and Acoustic doppler velocimeter(ADV) have been developed to measure velocities in in a fluid flow.

(a) The speed of the wave is the distance divided by the required time. Thus,

$$v = \dfrac{853 \space seats} {39 \space s} = 21.87 \space seat/s \approx 22 \space seats/s.$$

(b) The width $$w$$ is equal to the distance the wave has moved during the average time required by a spectator to stand and then sit. Thus,

$$w = vt = (21.87\space  seats/s) (1.8 \space s) \approx 39$$ seats

The speed of each individual wave is

$$v=\sqrt{\dfrac{\tau}{\mu}}=\sqrt{\dfrac{40 \mathrm{N}}{(0.125 \mathrm{kg})(2.25 \mathrm{m})}}=26.83 \mathrm{m} / \mathrm{s}$$

The average rate at which energy is transmitted from one side is

$$P_{\mathrm{avgl}}=\dfrac{1}{2} \mu v \omega^{2} y_{m}^{2}=\dfrac{1}{2}\left(\dfrac{0.125 \mathrm{kg}}{2.25 \mathrm{m}}\right)(26.83 \mathrm{m} / \mathrm{s})(2 \pi \times 120 \mathrm{Hz})^{2}\left(5.0 \times 10^{-3} \mathrm{m}\right)^{2}=10.6 \mathrm{W}$$

From both sides, $$ P_{\text {avg }}=2 P_{\text {wgl }}=2(10.6 \mathrm{W})=21.2 \mathrm{W} $$

The rate of change of kinetic energy from one side is given by

$$\dfrac{d K_{1}}{d t}=\dfrac{1}{2} \mu v \omega^{2} y_{m}^{2} \cos ^{2}(k x-\omega t)$$

Integrating over one period for both sides, we obtain

$$K =\int\left(2 \dfrac{d K_{1}}{d t}\right) d t=\mu v \omega^{2} y_{m}^{2} \int_{0}^{T} \cos ^{2}(k x-\omega t) d t=\dfrac{T}{2} \mu v \omega^{2} y_{m}^{2}=\dfrac{P_{\mathrm{avg}}}{2 f}$$

$$K=\dfrac{21.2 \mathrm{W}}{2(120 \mathrm{Hz})}=8.83 \times 10^{-2} \mathrm{J}$$

$$ u $$ satisfying $$ u / v $$ < 1 . $$ 

The Doppler shift (with $$ u=-0.950 \mathrm{m} / \mathrm{s} $$ ) leads to

$$\left.f_{\text {beat }}=\left|f_{r}-f_{s}\right| \approx \dfrac{2|u|}{v} f_{s}=\dfrac{2(0.95 \mathrm{m} / \mathrm{s})(28.0 \mathrm{kHz})}{343 \mathrm{m} / \mathrm{s}}\right)$$

$$=155 \mathrm{Hz}$$

(b) $$y=10 \cos [(252-250) \ \ \pi t] \ cos [(252+250) \ x$$]= $$10\cos500x$$

here y doesn't depend on time, so wave is stationary

(c) It is known that $$y=A\cos (\omega t+kx)$$ represent a wave travelling in positive x direction when $$k<0$$ and represent a wave travelling in negative x direction when $$k>0$$
in $$4 \sin \bigg(  5x-\dfrac{t}{2} \bigg) + 3 \cos \bigg( 5x-\dfrac{t}{2} \bigg)$$; 

so, $$f_A=480+5=485$$ Hz or $$480-5=475$$ Hz

On loading fork A its frequency decreases. But the beat decreases to 3 beats per sec. 

So, $$f_A$$ is greater than $$f_B$$ . Thus, $$f_A=485 $$ Hz

given that the , beat frequency is 4/s ,

therefore , 

               $$n\pm256=4$$ ,

or            $$n=256\pm4=252$$ or $$262Hz$$ ,

   if the frequency of first fork is $$260Hz$$ , on loading with wax , its frequency decreases and difference in frequency (beat frequency) should decrease , but it is not happening , so $$260Hz$$ is not the frequency of first fork .

    if the frequency of first fork is $$252Hz$$ , on loading with wax , its frequency decreases and difference in frequency (beat frequency) should increase , but it is now happening as the beat frequency is now $$6/s$$ , so $$252Hz$$ is the original frequency of first fork .

   

Beat frequency will be observed at the source due to the presence of two frequencies there. one is its own original frequency and other will be the echo which will come after striking the wall.

Let there be an imaginary observer $$\left(0_1\right)$$ at the wall frequency observed by $$0_1 $$ (due to doppler effect ) $$=\left( \dfrac { { V }_{ sound } }{ { V }_{ sound }-{ V }_{ source } }  \right) { f }_{ 0 }$$

$$=\left(\dfrac{330}{330-5}\right) \left(256\right) \simeq 260$$

Now, the sound after striking the wall will go to source (echo)

Frequency of echo an observed by the source (due to doppler effect ) $$=\left( \dfrac { { V }_{ sound }+5 }{ { V }_{ sound } }  \right) \left(260\right)$$

                                                                                                                      $$=\left( \dfrac{330+5}{330}\right) \left(260\right)\simeq  264$$

$$\Rightarrow$$ Beat frequency $$=\left| { f }_{ echo }-{ f }_{ 0 } \right| =8$$

                                $$=8$$ beat per second.

Hence, the answer is $$8$$ beat per second.

Speed of sound in air$$ V = 332 m/s$$, 

 The speed of the observer, $$u = 3.0 m/s$$. 

 The frequency of the sources $$f_0 = 256 Hz.$$

  The sources are stationary and observer moving, hence the apparent frequency heard by the observer from the front tuning fork (approaching) 

$$f'=\dfrac{V+u}{V}\times f_0=\dfrac{(332+3)}{332} \times 256=258\ Hz$$

 and the apparent frequency heard by the observer from the receding source 

Hence the beat frequency observed by the listener is,

$$=f'-f''$$

$$=258-246$$

$$=12\ Hz$$

We have to use the concept of Doppler's effect here

$$v=\dfrac{300m}{60s}=5\,m/s$$

$$f_o=\dfrac{20}{60}=\dfrac{1}{3}/sec$$

$$\lambda=\dfrac{v}{f_o}=15\,m/s$$

$$f=f_o(\dfrac{v-v_r}{v-v_s})$$

Figure (9.23) shows two waves superposition at the same place as per the time and formation of beats. For convenience the frequency of one wave is taken 5 and that of the other is taken 4. The wave with frequency 5 is shown by dotted line and wave with frequency 4 is shown by a straight line. Vertical axis shows displacement of particles from the mean position and horizontal axis shows time. Wave formed by the superposition of both is shown by thick black line. At instant A, both the waves are in the same phase and hence resultant displacement will be maximum and intensity of sound will be highest. After small time difference B, both the waves will be in opposite phase and hence will deduct the effect of each other and will make zero displacement and zero sound. Again at instant C, both the waves will be in same phase with maximum displacement and loud sound. Similarly, according to time at any place sound will be maximum or minimum. Hence, we can say as conclusion; 

"Superposition of two slightly different frequencies waves produce sometimes high or low intensity sounds, this harmonic cycle is called beats."

Phase difference of two high intensity sounds or two low intensity sounds is called time of beat, meaning in this time one beat is produced. 

From the above figure one more conclusion is drawn that if first wave produces 5 vibrations in 1 second and second wave produces 4 vibrations in 1 second then in 1 second 1 beat is heard meaning frequency of beats is equal to the difference of the frequencies of two waves.