Combining all of these we get It can be shown that 2
f r 1 l A
frequency of a stretched string: k 1
f f 1
2lA
T m
frequency, l length, T tension,m mass per unit length SAMPLE PROBLEM 11D
A wire of length 7.5 cm and mass 3.75 g is stretched between two points with a tension of 45 N. Calculate the fundamental frequency of vibration of the string.
SAMPLE ANSWER 11D Using the formula
f 1 2lA
T m
The fundamental frequency is 200 Hz. 1
f 1 2l A
T m
2(7.5 102)A0.05 1 1 45
T m
.This implies
f k l A
T m
. . This gives us the following equation for the fundamental
FUNDAMENTAL FREQUENCY OF A STRETCHED STRING
, making sure all values are in their standard unit:
m mass per unit length mass length 3.75 103
0.15
7.5 102 0.05 kg m1 2900 1
0.15 (30) 200
EXERCISE 11.2 HARMONICS
Assume the speed of sound in air is 340 m s1 for all these questions. Q1 A church organ consists of a series of pipes that are open at both ends, the longest being 2.4 m.What is the frequency of: (i) the fundamental note (ii) the second harmonic (iii) the third harmonic.
Q2 What is the frequency of the fundamental note produced by a clarinet of length 29.5 cm, given that it is essentially a pipe that is closed at one end and open at the other? What is the frequency of the next lowest harmonic present in the clarinet?
Q3 The steel piano wire has a mass per unit length of 0.0835 kg m1. Calculate frequency of the funda- mental note produced by a wire of length 1.4 m under a tension of 700 N.
Q4 A guitar’s E-string has length 66 cm and is stretched to a tension of 83 N. It vibrates at a fundamental frequency of 329.77 Hz. Determine the mass per unit.
