Question

A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be _________ cm. (Take speed of sound in air as 340 ms\(^{-1}\))

Hint:

Remember the fundamental frequency formulas for pipes: \(f_1 = \frac{v}{4L}\) for a closed pipe and \(f_1 = \frac{v}{2L}\) for an open pipe (open at both ends). The shortest length always corresponds to the fundamental frequency (n=1). Always check the units required for the final answer (m vs cm).

Answer 1

Correct Answer: 34

Step 1: Understanding the Question:
We need to find the length of a closed organ pipe that will produce its fundamental frequency (since we want the shortest pipe) in resonance with a tuning fork of a given frequency.
Step 2: Key Formula or Approach:
A closed organ pipe (closed at one end, open at the other) supports standing waves where the closed end is a node and the open end is an antinode.
The resonant frequencies are given by the formula:
\[ f_n = n \frac{v}{4L} \quad \text{where } n = 1, 3, 5, ... \text{ (odd integers)} \] - \(f_n\) is the frequency of the n-th harmonic.
- \(v\) is the speed of sound.
- \(L\) is the length of the pipe.
The fundamental frequency (for the shortest pipe, n=1) is \( f_1 = \frac{v}{4L} \).
Step 3: Detailed Explanation:
For the pipe to resonate with the tuning fork, its fundamental frequency must match the tuning fork's frequency.
Given values:
- Frequency of tuning fork, \( f = 250 \) Hz.
- Speed of sound in air, \( v = 340 \) m/s.
Set the fundamental frequency of the pipe equal to the tuning fork's frequency:
\[ f_1 = f = 250 \, \text{Hz} \] Using the formula for the fundamental frequency:
\[ 250 = \frac{340}{4L} \] Now, solve for the length L.
\[ 4L = \frac{340}{250} \] \[ 4L = \frac{34}{25} = 1.36 \, \text{m} \] \[ L = \frac{1.36}{4} = 0.34 \, \text{m} \] The question asks for the length in centimeters.
\[ L = 0.34 \, \text{m} \times 100 \, \text{cm/m} = 34 \, \text{cm} \] Step 4: Final Answer:
The length of the shortest closed organ pipe is 34 cm.