In a resonance tube experiment, a tube of fixed length is closed at one end and several tunning forks of increasing frequency used to obtain resonance at the open end. If the tunning fork with the lowest frequency which gave resonance had a frequency f1 and the next tunning fork to give resonance had a frequency f2, find the ratio f2:f1
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
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Step 1: Determine the formula for resonant frequencies in a closed-end tube.
For a resonance tube closed at one end and open at the other, resonance occurs when the length of the tube $L$ is an odd multiple of a quarter wavelength. The resonant frequencies $f_n$ are given by:
$$f_n = \frac{nv}{4L}$$
where $n = 1, 3, 5, \dots$ (odd integers) and $v$ is the speed of sound in air.
Step 2: Find the lowest resonant frequency ($f_1$).
The lowest frequency corresponds to $n=1$.
$$f_1 = \frac{1 \cdot v}{4L} = \frac{v}{4L}$$
Step 3: Find the next resonant frequency ($f_2$).
The next resonant frequency corresponds to the next odd integer, which is $n=3$.
$$f_2 = \frac{3 \cdot v}{4L}$$
Step 4: Calculate the ratio $f_2:f_1$.
To find the ratio $\frac{f_2}{f_1}$, we divide the expression for $f_2$ by the expression for $f_1$:
$$\frac{f_2}{f_1} = \frac{\frac{3v}{4L}}{\frac{v}{4L}}$$
$$\frac{f_2}{f_1} = \frac{3v}{4L} \times \frac{4L}{v}$$
$$\frac{f_2}{f_1} = 3$$
So, the ratio $f_2:f_1$ is $3:1$.
The ratio $f_2:f_1$ is $\boxed{\text{3:1}}$.
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