Question 362: Waves

Question

When two tuning forks are sounded together, 5 beats per second are heard. One of the forks is in unison with $$0.97 \mathrm{~m}$$ length of sonometer wire and the other is in unison with $$0.96 \mathrm{~m}$$ length of the same wire. The frequencies of the two tuning forks are

Accepted Answer

Let, $$f_1$$ and $$f_2$$ be the frequency of the two turning forks, given
$$f_1-f_2=5$$ ..... (i)
Length of wire in sonometer is
$$L_1=0.96 \mathrm{~m} 	ext { and } L_2=0.97 \mathrm{~m}$$
Frequency of vibration of the wire is given by
$$f=\frac{1}{2 l} \sqrt{T / m}$$
According to the question,
$$\begin{aligned}
& \frac{f_1}{f_2}=\frac{l_2}{l_1}=\frac{0.97}{0.96}>1 \
	herefore \quad & f_1>f_2
\end{aligned}$$
Now, from Eq. (i)
$$\begin{array}{ll}
\Rightarrow & f_1-f_2=5 \
\Rightarrow & \frac{0.97}{0.96} f_2-f_2=5 \
\Rightarrow & {\left[\frac{0.97}{0.96}-1ight] f_2=5} \
\Rightarrow & {\left[\frac{0.97-0.96}{0.96}ight] f_2=5} \
\Rightarrow & 0.01 f_2=5 	imes 0.96 \
\Rightarrow & 0.01 f_2=4.8 \
\Rightarrow & f_2=480 \mathrm{~Hz} \
\Rightarrow & f_1=5+f_2=485 \mathrm{~Hz}
\end{array}$$

Answer

$$383 \mathrm{~Hz}, 388 \mathrm{~Hz}$$

Answer

$$475 \mathrm{~Hz}, 480 \mathrm{~Hz}$$

Answer

$$388 \mathrm{~Hz}, 39 z \mathrm{~Hz}$$

Answer

$$480 \mathrm{~Hz}, 485 \mathrm{~Hz}$$

Waves

When two tuning forks are sounded together, 5 beats per second are heard. One of the forks is in unison with $0.97 \mathrm{~m} length of sonometer wire and the other is in unison with 0.96 \mathrm{~m}$ length of the same wire. The frequencies of the two tuning forks are

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