Question 304: Waves

Question

The frequency of the third overtone of a pipe of length ' $L_{\mathrm{c}}$ ', closed at one end is same as the frequency of the sixth overtone of a pipe of length ' $L_0$ ', open at both ends. Then the ratio $\mathrm{L}_{\mathrm{c}}: \mathrm{L}_0$ is

Accepted Answer

Let's break down the problem step by step:
For a pipe closed at one end, only odd harmonics are present. The frequency of the nth harmonic is given by
$$f_n = \frac{nv}{4L_{\mathrm{c}}},$$
where $$n =1, 3, 5, \dots$$.
The fundamental frequency (0th overtone) corresponds to $$n=1$$.
The first overtone corresponds to $$n=3$$.
The second overtone corresponds to $$n=5$$.
The third overtone corresponds to $$n=7$$.
Therefore, the frequency of the third overtone is
$$f_{\mathrm{closed}} = \frac{7v}{4L_{\mathrm{c}}}.$$
For an open pipe, all harmonics are present. The frequency of the nth harmonic is given by
$$f_n = \frac{nv}{2L_0},$$
where $$n = 1, 2, 3, \dots$$.
The fundamental frequency (0th overtone) corresponds to $$n=1$$.
The first overtone corresponds to $$n=2$$.
The second overtone corresponds to $$n=3$$.
The third overtone corresponds to $$n=4$$.
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The sixth overtone corresponds to $$n=7$$.
So, the frequency of the sixth overtone is
$$f_{\mathrm{open}} = \frac{7v}{2L_0}.$$
According to the problem, these two frequencies are equal:
$$\frac{7v}{4L_{\mathrm{c}}} = \frac{7v}{2L_0}.$$
We can cancel $$7v$$ from both sides, simplifying the equation to:
$$\frac{1}{4L_{\mathrm{c}}} = \frac{1}{2L_0}.$$
Cross-multiplying gives:
$$2L_0 = 4L_{\mathrm{c}},$$
which simplifies to:
$$L_0 = 2L_{\mathrm{c}}.$$
Rearranging the ratio $$L_{\mathrm{c}}:L_0$$, we get:
$$L_{\mathrm{c}} : L_0 = 1 : 2.$$
Thus, the correct answer is Option B: $$1:2.$$

Answer

$1: 4$

Answer

$1: 2$

Answer

$2: 1$

Answer

$4: 1$

Waves

The frequency of the third overtone of a pipe of length ' L_{\mathrm{c}} ', closed at one end is same as the frequency of the sixth overtone of a pipe of length ' L_0 ', open at both ends. Then the ratio \mathrm{L}_{\mathrm{c}}: \mathrm{L}_0 is

Choose the correct answer: