The intensity ratio of the two waves is 1 : 16. The ratio of their amplitudes is

(1) 1 : 16

(2) 1 : 4

(3) 4 : 1

(4) 2 : 1

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The superposing waves are represented by the following equations : ${y}_{1}=5\mathrm{sin}2\pi \left(10\text{\hspace{0.17em}}t-0.1x\right)$, ${y}_{2}=10\mathrm{sin}2\pi \left(20\text{\hspace{0.17em}}t-0.2x\right)$ Ratio of intensities $\frac{{I}_{\mathrm{max}}}{{I}_{\mathrm{min}}}$ will be :

(1) 1

(2) 9

(3) 4

(4) 16

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The displacement of a particle is given by $x=3\mathrm{sin}\left(5\pi \text{\hspace{0.17em}}t\right)+4\mathrm{cos}\left(5\pi \text{\hspace{0.17em}}t\right)$. The amplitude of the particle is :

(1) 3

(2) 4

(3) 5

(4) 7

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The two interfering waves have intensities in the ratio 9 : 4. The ratio of intensities of maxima and minima in the interference pattern will be :

(1) 1 : 25

(2) 25 : 1

(3) 9 : 4

(4) 4 : 9

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

If the ratio of amplitude of two waves is 4 : 3. Then the ratio of maximum and minimum intensity will be :

(1) 16 : 18

(2) 18 : 16

(3) 49 : 1

(4) 1 : 49

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

Equation of motion in the same direction is given by ${y}_{1}=A\mathrm{sin}\left(\omega t-kx\right)$, ${y}_{2}=A\mathrm{sin}\left(\omega t-kx-\theta \right)$. The amplitude of the medium particle will be

(1) $2A\mathrm{cos}\frac{\theta }{2}$

(2) $2A\mathrm{cos}\theta$

(3) $\sqrt{2}A\mathrm{cos}\frac{\theta }{2}$

(4) $1.2f,\text{\hspace{0.17em}}1.2\lambda$

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The displacement of the interfering light waves are ${y}_{1}=4\mathrm{sin}\omega \text{\hspace{0.17em}}t$ and ${y}_{2}=3\mathrm{sin}\left(\omega \text{\hspace{0.17em}}t+\frac{\pi }{2}\right)$. What is the amplitude of the resultant wave :

(1) 5

(2) 7

(3) 1

(4) 0

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

Two waves are represented by ${y}_{1}=a\mathrm{sin}\left(\omega \text{\hspace{0.17em}}t+\frac{\pi }{6}\right)$ and ${y}_{2}=a\mathrm{cos}\omega \text{\hspace{0.17em}}t$. What will be their resultant amplitude :

(1) a

(2) $\sqrt{2}\text{\hspace{0.17em}}a$

(3) $\sqrt{3}\text{\hspace{0.17em}}a$

(4) 2a

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The amplitude of a wave represented by displacement equation $y=\frac{1}{\sqrt{a}}\mathrm{sin}\omega t±\frac{1}{\sqrt{b}}\mathrm{cos}\omega t$ will be

(1) $\frac{a+b}{ab}$

(2) $\frac{\sqrt{a}+\sqrt{b}}{ab}$

(3) $\frac{\sqrt{a}±\sqrt{b}}{ab}$

(4) $\sqrt{\frac{a+b}{ab}}$

Concept Questions :-

Wave motion
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

Two tuning forks when sounded together produced 4 beats/sec. The frequency of one fork is 256 Hz. The number of beats heard increases when the fork of frequency 256 Hz is loaded with wax. The frequency of the other fork is(in Hz) :

(1) 504

(2) 520

(3) 260

(4) 252

Concept Questions :-

Beats