NEET Practice Questions (MCQs) with Answers & Solutions

Wave velocity $v=\frac{\lambda }{T}=\frac{\omega \lambda }{2\mathrm{\pi }}$

Maximum particle velocity ${\left(v_{max}\right)}_p=A\omega$

Given,            $v={\left(v_{max}\right)}_p$

                    $\frac{\omega \lambda }{2\pi }=A\omega$

$\Rightarrow \lambda =2\pi A$

Suppose $n_p$ = frequency of piano = ?

($n_p\propto \sqrt{T}$)

$n_f$ = frequency of tuning fork = 512Hz

x = Beat frequency = 4 beats/s, which is decreasing (4$\to$2) after changing the tension of piano wire.Also, tension of piano wire is increasing so $n_p\uparrow$

Hence, $n_p\uparrow -n_f=x\downarrow \to wrong$

          $n_f-n_p\uparrow =x\downarrow \to correct$

$\Rightarrow n_p=n_f-x=512-4=508 Hz$

Key Idea  Find the parameters and put them in the general wave equation.

                 Here,        A=2 cm

                     direction = +ve direction

                                $\mathrm{v}=128 {\mathrm{ms}}^{-1}$

     $$\begin{gathered} \mathrm{and} 5\mathrm{\lambda }=4 \\ \mathrm{Now}, \mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}=\frac{2\mathrm{\pi }\times 5}{4}=7.85 \\ \mathrm{and} \mathrm{v}=\frac{\mathrm{\omega }}{\mathrm{k}}=128 {\mathrm{ms}}^{-1} \\ \Rightarrow \mathrm{\omega }=\mathrm{v}\times \mathrm{k}=128\times 7.85 \\ =1005 \\ \mathrm{As}, \mathrm{y}=\mathrm{A} \sin (\mathrm{kx}-\mathrm{\omega t}) \\ \therefore \mathrm{y}=2 \sin (7.85 \mathrm{x}-1005 t ) \\ =(0.02) \mathrm{m} \sin (7.85 \mathrm{x}-1005 t ) \end{gathered}$$

The number of beats will be the difference of frequencies of the two strings.

Frequency of first string $f_1=\frac{1}{2l_1}\sqrt{\frac{T}{m}}$

             $=\frac{1}{2\times 51.6\times {10}^{-2}}\sqrt{\frac{20}{{10}^{-3}}}$

            =137.03 Hz           

Similarly, frequency of second string

             =$\frac{1}{2\times 49.1\times {10}^{-2}}\sqrt{\frac{20}{{10}^{-3}}}$

           =144.01

Number of beats=$f_2-f_1=144-137$

                      =7 beats

Resultant intensity of two periodic waves is given by 

           $I=I_1+I_2+2\sqrt{I_1{I}_2} \cos \delta$

where $\delta$ is the phase differences between the waves.  

for maximum intensity,

   $\delta =2n\pi ; n= 0,1,2....etc.$

Therefore, for zero order maxima, cos $\delta =1$

$I_{max}=I_1+I_2+2\sqrt{I_1{I}_2}={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2$

for minimum intensity , $\delta =\left(2n-1\right) \pi ;$

        n=1,2,.....etc

 Therefore, for Ist order minma, cos $\delta =-1$

      $$\begin{gathered} I_{min }=I_1+I_2-2 \sqrt{I_1{I}_2} \\ = {\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2 \\ Therefore , \\ {I}_{max}+I_{min}={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2+{\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2 \\ =2\left(I_1+I_2\right) \end{gathered}$$

Phase difference =$\frac{2\pi }{\lambda }\times path difference$

path difference between two points, $∆x=15-10=5m$

Time period, T=0.05 s

$$\begin{gathered} \Rightarrow frequency v=\frac{1}{T}=\frac{1}{0.05}=20Hz \\ Velocity,v=300 m/s \\ \therefore Wavelength, \lambda =\frac{v}{v}=\frac{300}{20}=15m \\ Hence,phase difference \\ ∆\varphi =\frac{2\pi }{\lambda }\times ∆x \\ =\frac{2\pi }{15}\times 5=\frac{2\pi }{3} \end{gathered}$$

The sign between two terms in argument of sine will define its direction. 

Writing the given wave equation 

$y=0.25 \sin \left(10\pi x-2\pi t\right) ...\left(i\right)$

The minus (-) between $\left(10\pi x\right) and \left(2\pi t\right)$ implies that the wave is travelling along positive x direction. 

Now comparing Eq. (i) with standard wave equation 

$$\begin{gathered} y=a \sin \left(kx-wt\right) ...\left(ii\right) \\ we have \\ a=0.25 m, w= 2\pi , k=10\pi m \\ \therefore \frac{2\pi }{T}=2\pi \\ \Rightarrow f=1 Hz \\ Also, \lambda =\frac{2\pi }{k}=\frac{2\pi }{10\pi }=0.2 m \end{gathered}$$

Therefore, the wave is travelling along +ve x direction with frequency 1 Hz and wavelength 0.2 m

$v=n\lambda =2\times 5=10$ cm/sec

$v=n\lambda$ ⇒ $\lambda =\frac{v}{n}=\frac{330}{256}=1.29m$