NEET Practice Questions (MCQs) with Answers & Solutions

$$\begin{gathered} Given, \\ {y}_1=a\sin \omega t \\ {y}_2=b\cos \omega t=b\sin (\omega t+\pi /2) \\ The resul\tan t displacement is given by; \\ y=y_1+y_2=\sqrt{a^2 + b^2 +2ab\cos \frac{\mathrm{\pi }}{2}}\sin (\omega t+\phi )=\sqrt{a^2 + b^2 }\sin (\omega t+\phi ) \\ Hence, the motion of superimposed wave is simple harmonic with amplitude \sqrt{a^2 + b^2 }. \end{gathered}$$

For a particle executing SHM, we have maximum acceleration,
α=Aω²   ...(i)

Where, A is max. amplitude and ω is angular velocity of a particle

Max. velocity,β=Aω ...(ii)

Dividing Eq. (i) by Eq. (ii), we get

α/β=Aω²/Aω

=>α/β=ω=2π/T

i.e. T=2πβ/α

Thus,its time period of vibration, T=2πβ/α

Given,

Damping force $\propto$ velocity

              $F\propto kv$ 

             $F=kv$

            $k=\frac{F}{v}$

Unit of k =$\frac{unit of F}{unit of v}$

            $=\frac{kg m{s}^{-2}}{m{s}^{-1}}=kg{s}^{-1}$

Answer given is correct

$$\begin{gathered} x=a{\sin }^2\omega t \\ u\sin g, \cos 2\omega t =1-2{\sin }^2\omega t \\ {\sin }^2\omega t =\frac{1-\cos 2\omega t}{2} \\ So, x =\frac{a}{2}-\frac{a\cos 2\omega t}{2} \end{gathered}$$

We have reduced the question to a single function

So, T=$\frac{\mathrm{\pi }}{\omega }$ and f= $\frac{\omega }{\mathrm{\pi }}$

Time period of spring pendulum, T= $2\mathrm{\pi }\sqrt{\frac{\mathrm{M}}{\mathrm{k}}}.$

If now mass in doubled $T^{\text{'}}=2\mathrm{\pi }\sqrt{\frac{2\mathrm{M}}{\mathrm{k}}}=\sqrt{2}\mathrm{T}$

$$\begin{gathered} \mathrm{v}=\frac{dy}{dt}=A\omega \mathit{ }cos\mathit{ }\omega t=A\omega \sqrt{1-{\sin }^2 \omega t} \\ =\omega \sqrt{{\mathrm{A}}^2-{\mathrm{y}}^2} \\ \mathrm{Here}, \mathrm{y}=\frac{\mathrm{a}}{2} \\ \therefore \mathrm{v}=\mathrm{\omega }\sqrt{{\mathrm{a}}^2-\frac{{\mathrm{a}}^2}{4}}=\mathrm{\omega }\sqrt{\frac{3{\mathrm{a}}^2}{4}}=\frac{2\mathrm{\pi }}{\mathrm{T}}\frac{\mathrm{a}\sqrt{3}}{2}=\frac{\mathrm{\pi a}\sqrt{3}}{\mathrm{T}} \end{gathered}$$

Acceleration $\propto -\left(displacement\right).$

           $A\propto -y$

          $A=-{\omega }^{2}y$

         $A=-\frac{k}{m}y$

      $A=-k\text{'}y$         

Here, y=x+a

$\therefore$acceleration =-k(x+a)

Acceleration of simple harmonic motion is 

         $a_{max}=-{\omega }^{2}A$

or $\frac{{\left(a_{max}\right)}_1}{{\left(a_{max}\right)}_2}=\frac{{\omega }_1^2}{{\omega }_2^2}$                               (as A remains same)

or $\frac{{\left(a_{max}\right)}_1}{{\left(a_{max}\right)}_2}=\frac{{\left(100\right)}^2}{{\left(1000\right)}^2}={\left(\frac{1}{10}\right)}^2=1:{10}^2$

Velocity is the time derivation of displacement. Writing the given equation of a point performing SHM 

        x= a sin $\left(\omega t+\frac{\pi }{6}\right)$     ..(i)

Differentiating Eq. (i),w.r.t. time, we obtain 

            v=$\frac{dx}{dt}=a \omega \cos \left(\omega t+\frac{\pi }{6}\right)$

     It is given that v=$\frac{a\omega }{2}$,so that 

            $$\begin{gathered} \frac{a\omega }{2}=a\omega \cos \left(\omega t+\frac{\pi }{6}\right) \\ or \frac{1}{2}=\cos \left(\omega t+\frac{\pi }{6}\right) \\ or \cos \frac{\pi }{3}= \cos \left(\omega t+\frac{\pi }{6}\right) \\ or \omega t+\frac{\pi }{6}=\frac{\pi }{3} \\ \Rightarrow \omega t=\frac{\pi }{6} \\ or t=\frac{\pi }{6\omega }=\frac{\pi \times T}{6\times 2\pi }=\frac{T}{12} \end{gathered}$$

Thus, at $\frac{T}{12}$ velocity of the point will be equal to half of its  maximum velocity.