Question 338: Waves

Question

The end correction of resonance tube is 1 cm. If the shortest length resonating with a tuning fork is 15 cm , the next resonating length will be

Accepted Answer

Let the shortest resonating length be denoted as $ l_1 = 15.0\, 	ext{cm} $.
The end correction is $ e = 1.0\, 	ext{cm} $.
We can express the shortest resonating length with the tuning fork as:
$ l_1 + e = \frac{\lambda}{4} $
where $\lambda$ is the wavelength.
For the next higher resonating length, it will correspond to $ \frac{3\lambda}{4} $, so:
$ l_2 + e = \frac{3\lambda}{4} $
To find the ratio of these expressions:
$ \begin{align*} 	herefore & \frac{l_1 + e}{l_2 + e} = \frac{1}{3} \ 	herefore & \frac{15 + 1}{l_2 + 1} = \frac{1}{3} \ 	herefore & l_2 + 1 = 48 \ 	herefore & l_2 = 47\, 	ext{cm} \end{align*} $
Thus, the next resonating length is $ 47\, 	ext{cm} $.

Answer

35 cm

Answer

40 cm

Answer

47 cm

Answer

64 cm

Waves

The end correction of resonance tube is 1 cm. If the shortest length resonating with a tuning fork is 15 cm , the next resonating length will be

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