In a guitar, two strings \(A\) and \(B\) made of same material are slightly out of tune and produce beats of frequency \(6~\text{Hz}\). When tension in \(B\) is slightly decreased, the beat frequency increases to \(7~\text{Hz}\).  If the frequency of \(A\) is \(530~\text{Hz}\), the original frequency of \(B\) will be:

Three sound waves of equal amplitudes have frequencies of \((n-1),~n,\) and \((n+1).\) They superimpose to give beats. The number of beats produced per second will be:

A source of unknown frequency gives \(4\) beats/s when sounded with a source of known frequency of \(250~\text{Hz}\). The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency of \(513~\text{Hz}\). The unknown frequency will be:

Two sources of sound placed close to each other, are emitting progressive waves given by,
\(y_1=4\sin 600\pi t\) and \(y_2=5\sin 608\pi t\).
${}_{}$An observer located near these two sources of sound will hear:

Two identical piano wires, kept under the same tension T, have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to the occurrence of 6 beats/s when both the wires oscillate together would be:

1. 0.01

2. 0.02

3. 0.03

4. 0.04

A tuning fork of frequency \(512\) Hz makes \(4\) beats/s with the vibrating string of a piano. The beat frequency decreases to \(2\) beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was:
1. \(510\) Hz
2. \(514\) Hz
3. \(516\) Hz
4. \(508\) Hz

Each of the two strings of lengths 51.6 cm and 49.1 cm is tensioned separately by 20 N of force. The mass per unit length of both strings is the same and equals 1 g/m. When both the strings vibrate simultaneously, the number of beats is:

Two sound waves with wavelengths 5.0 m and 5.5 m, respectively, propagate in gas with a velocity of 330 m/s. How many beats per second can we expect?

1. 12

2. 0

3. 1

4. 6

Two vibrating tuning forks produce progressive waves given by \(Y_1 = 4 ~\mathrm{sin}~500 \pi \mathrm{t}\) and \(Y_2 = 2 ~\mathrm{sin}~506 \pi \mathrm{t}\). The number of beats produced per minute is:

Two stationary sources exist, each emitting waves of wavelength λ. If an observer moves from one source to the other with velocity u, then the number of beats heard by him is equal to:

1. $\frac{2u}{\lambda }$

2. $\frac{u}{\lambda }$

3. $\sqrt{\mu \lambda }$

4. $\frac{\mu }{2\lambda }$