This document contains two multiple choice physics questions about beat frequency. Question 1 asks about the unknown frequency of a tuning fork that produces different beat frequencies when sounded with 238 Hz and 250 Hz tuning forks. The unknown frequency is calculated to be 248 Hz. Question 2 asks about the length of a pipe cut off based on the beat frequency produced when sounded with an identical pipe of known 325 Hz frequency. The calculation determines 1.16 cm was cut off from the open end of the second pipe.

1. Physics 101 Learning Object
Beat Frequency Multiple Choice Questions
Question 1: A tuning fork of unknown frequency produces 50 beats in 5.0s when sounded with
another 238 Hz tuning fork and 10 beats in 5.0s when sounded with a 250 Hz tuning fork. What
is the unknown frequency?
A. 244 Hz
B. 248 Hz
C. 240 Hz
D. 246 Hz
Question 2: A pipe that is closed at one end and open at the other end has a fundamental
frequency of 325 Hz. Another pipe that is identical to the first one has a part of its' open end cut
off. Now when both pipes vibrate at their fundamental frequencies, a beat frequency of 15 Hz is
heard. Given that the speed of sound is 343 m/s, how much of the second pipe was cut off at it's
open end?
A. 2.30 cm
B. 1.53 cm
C. 1.16 cm
D. 2.27 cm

2. Solutions
Question 1: A tuning fork of unknown frequency produces 50 beats in 5.0s when sounded with
another 238 Hz tuning fork and 10 beats in 5.0s when sounded with a 250 Hz tuning fork. What
is the unknown frequency?
A. 244 Hz
B. 248 Hz
C. 240 Hz
D. 246 Hz
Beat frequency = |f2 - f1|
First tuning fork: 50 beats in 5.0s = 50/5= 10 beats per second. Therefore the beat frequency is
10. The tuning fork is 238 Hz, so the unknown is 238 ± 10 = 228 or 248.
Second tuning fork: 10 beats in 5.0s = 10/5 = 2 beats per second. Thefore the beat frequency is
2. The tuning fork is 250 Hz, so the unknown is 250 ± 2 = 248 or 252.
Therfore the unknown frequency is 248 Hz.
Question 2: A pipe that is closed at one end and open at the other end has a fundamental
frequency of 325 Hz. Another pipe that is identical to the first one has a part of its' open end cut
off. Now when both pipes vibrate at their fundamental frequencies, a beat frequency of 15 Hz is
heard. Given that the speed of sound is 343 m/s, how much of the second pipe was cut off at it's
open end?
A. 2.30 cm
B. 1.53 cm
C. 1.16 cm
D. 2.27 cm
For a pipe that has one end closed and one end open: f = V/4L, therefore L = V/4f.
Normal pipe: L = 343/(4*325) = 0.2638 m
Beat frequency = |f2 - f1|
15 = |f2 - 325|

3. f2 = 340 (the frequency increases as the length of the pipe is shortened)
Pipe with part of open end cut off: L= 343/(4*340) = 0.2522 m
Therefore the amount cut off = 0.2638 - 0.2522 = 0.0116 m = 1.16cm