Two waves are represented by $y_1=a\sin \left(\omega t+\frac{{\displaystyle \pi }}{{\displaystyle 6}}\right)$ and $y_2=a\cos \omega t$. What will be their resultant amplitude :

(1) a

(2) $\sqrt{2}a$

(3) $\sqrt{3}a$

(4) 2a

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

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

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

(3) $\frac{\sqrt{a}\pm \sqrt{b}}{ab}$

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

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

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 :

(1) 257

(2) 258

(3) 256

(4) 254

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 :

(1) 250

(2) 252

(3) 260

(4) 262

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 :

(1) 1/4 sec

(2) 1/2 sec

(3) 1 sec

(4) 2 sec

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 :

1. 334

2. 339

3. 343

4. 347

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 :

(1) Zero

(2) One

(3) Four

(4) Eight