Question 11:
easyFor production of beats the two sources must have:
(1992)
Beats are the periodic variations in intensity of sound when two sound waves of slightly different frequencies superimpose. Thus, they must primarily have different frequencies.
Question 12:
easyTwo sound waves with wavelengths $5.0 \text{ m}$ and $5.5 \text{ m}$ respectively, each propagate in a gas with velocity $330 \text{ m/s}$. We expect the following number of beats per second:
(2006)
The frequencies are $f_1 = \frac{v}{\lambda_1} = \frac{330}{5.0} = 66 \text{ Hz}$ and $f_2 = \frac{v}{\lambda_2} = \frac{330}{5.5} = 60 \text{ Hz}$. Number of beats = $f_1 - f_2 = 66 - 60 = 6$.
Question 13:
easyTwo 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:
(2005)
Angular frequencies are $\omega_1 = 500 \pi$ and $\omega_2 = 506 \pi$, so $f_1 = 250 \text{ Hz}$ and $f_2 = 253 \text{ Hz}$. Beats per second = $253 - 250 = 3$. Beats per minute = $3 \times 60 = 180$.
Question 14:
easyA source of sound gives 5 beats per second, when sounded with another source of frequency $100 \text{ second}^{-1}$. The second harmonic of the source, together with a source of frequency $205 \text{ sec}^{-1}$ gives 5 beats per second. What is the frequency of the source?
(1995)
Frequency $f = 100 \pm 5 = 105$ or $95 \text{ Hz}$. Second harmonic $2f = 210$ or $190 \text{ Hz}$. Since $|2f - 205| = 5$, $2f$ must be $210$. Thus $$f = 105 \text{ second}^{-1}$$.
Question 15:
easyA source of frequency $v$ gives 5 beats/second when sounded with a source of frequency $200 \text{ Hz}$. The second harmonic of frequency $2v$ of source gives 10 beats/second when sounded with a source of frequency $420 \text{ Hz}$. The value of $v$ is:
(1994)
Frequency $v = 200 \pm 5 = 205$ or $195 \text{ Hz}$. Also $|2v - 420| = 10$. If $v = 205$, then $2v = 410$, and $|410 - 420| = 10$, which matches. Thus $v = 205 \text{ Hz}$.
Question 16:
easyWave has simple harmonic motion whose period is 4 seconds while another wave which also possesses simple harmonic motion has its period 3 sec. If both are combined, then the resultant wave will have the period equal to:
(1993)
The period of the resultant wave formed by the superposition of two periodic waves is the least common multiple (LCM) of their individual periods. $$\text{LCM}(4, 3) = 12 \text{ seconds}$$.