Two sound waves (expressed in CGS units) given by $y_1=0.3\sin \frac{2\pi }{\lambda }(vt-x)$ and $y_2=0.4\sin \frac{2\pi }{\lambda }(vt-x+\theta )$ interfere. The resultant amplitude at a place where the phase difference is π/2 will be :

If two waves having amplitudes 2A and A and same frequency and velocity, propagate in the same direction in the same phase, the resulting amplitude will be 

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

The superposing waves are represented by the following equations : $y_1=5\sin 2\pi (10t-0.1x)$, $y_2=10\sin 2\pi (20t-0.2x)$ Ratio of intensities $\frac{I_{\max }}{I_{\min }}$ will be :

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

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 :

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

Equation of motion in the same direction is given by $y_1=A\sin (\omega t-kx)$, $y_2=A\sin (\omega t-kx-\theta )$. The amplitude of the medium particle will be 

The displacement of the interfering light waves are $y_1=4\sin \omega t$ and $y_2=3\sin \left(\omega t+\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)$. What is the amplitude of the resultant wave :

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

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) :

A tuning fork sounded together with a tuning fork of frequency 256 Hz emits two beats. On loading the tuning fork of frequency 256 Hz with wax, the number of beats heard are 1 per second. The frequency of the other tuning fork is :

If two tuning forks A and B are sounded together, they produce 4 beats per second. A is then slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is 256. The frequency of B will be :

Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between successive maximum intensities will be :

When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is :

Tuning fork F₁ has a frequency of 256 Hz and it is observed to produce 6 beats/second with another tuning fork F₂. When F₂ is loaded with wax, it still produces 6 beats/second with F₁. The frequency of F₂ before loading was :

Two tuning forks, A and B, vibrating simultaneously produce 5 beats. The frequency of B is 512 Hz. It is seen that if one arm of A is filed a little, then the number of beats increases. The frequency of A in Hz will be :

Beats are produced by two waves given by $y_1=a\sin 2000\pi t$ and $y_2=a\sin 2008\pi t$. The number of beats heard per second is :

A tuning fork whose frequency as given by manufacturer is 512 Hz is being tested with an accurate oscillator. It is found that the fork produces a beat of 2 Hz when oscillator reads 514 Hz but produces a beat of 6 Hz when oscillator reads 510 Hz. The actual frequency of the fork is :

When a tuning fork A of unknown frequency is sounded with another tuning fork B of frequency 256 Hz, then 3 beats per second are observed. After that A is loaded with wax and sounded, the again 3 beats per second are observed. The frequency of the tuning fork A is :

When two sound waves are superimposed, beats are produced when they have :

A tuning fork A of frequency 200 Hz is sounded with fork B, the number of beats per second is 5. By putting some wax on A, the number of beats increases to 8. The frequency of fork B is :

Two tuning forks have frequencies 380 and 384 Hz respectively. When they are sounded together, they produce 4 beats. After hearing the maximum sound, how long will it take to hear the minimum sound?

A couple of tuning forks produces 2 beats in the time interval of 0.4 seconds. So the beat frequency is :

It is possible to hear beats from the two vibrating sources of frequency :

Two sound waves of wavelengths 5m and 6m formed 30 beats in 3 seconds. The velocity of sound is :

Two sound sources when sounded simultaneously produce four beats in 0.25 seconds. The difference in their frequencies must be :

Two strings X and Y of a sitar produce a beat frequency 4 Hz. When the tension of the string Y is slightly increased the beat frequency is found to be 2 Hz. If the frequency of X is 300 Hz, then the original frequency of Y was :

Two vibrating tuning forks produce progressive waves given by $Y_1=4\sin 500\pi t$ and $Y_2=2\sin 506\pi t.$ Number of beats produced per minute is :

The disc of a siren containing 60 holes rotates at a constant speed of 360 rpm. The emitted sound is in unison with a tuning fork of frequency :