This document discusses standing waves on strings. It defines standing waves as occurring from the interference of two waves moving in opposite directions with identical properties. Standing waves form nodes and antinodes due to constructive and destructive interference. The frequency of standing waves can be decreased by making the string longer, using a thicker string, or decreasing the tension. The document works through multiple choice and multi-part problems applying the formulas for string frequency and wave speed to calculate values for a specific string's harmonics and speed.

1. Investigation of Standing
Waves on Strings
Yu Han Zhang
#14050141

2. What are Standing Waves
Standing waves are produced by
repeated interference of 2 waves
with identical frequency,
wavelength, and amplitude while
moving in opposite directions.

3. Characteristics of Standing Waves
Nodes and anti-nodes are produced as a result of standing
waves.
Node: points on the string that have 0 amplitude and
remain at rest at all times. Nodes are caused by destructive
interference of the 2 waves
Anti-node: points that move with the maximum amplitude
(twice the original amplitude of each wave). Anti-nodes
are caused by constructive interference of the 2 waves

4. Patterns of Standing Waves
From the table, we can deduce some relationships between wavelength
and harmonic number; we can also find patterns of number of nodes and
anti-nodes and conclude how they are related to the harmonic number

5. Derivation of Formulas

6. Note that the harmonic
number m can only be
positive integers e.g.
1,2,3,4…

7. Multiple Choice 1
Now we have sufficient knowledge about standing waves, let us try
some questions. Start with some simple multiple choice questions
1. You can decrease the frequency of a standing wave on a string
by:
A. making the string longer
B. using a thicker string.
C. decreasing the tension.
D. all of the above.

8. Answer: D.All of the above
A. making the string longer - this works because as the
formula suggests: if length increases, frequency decreases
B. using a thicker string - this works because thicker
string means heavier mass, so linear mass density μ will
decrease; as a result, frequency decreases
C. decreasing tension - works because if T decreases, as
the formula suggest, frequency will decrease

9. Multiple Choice 2
A guitar string has a length of 60 cm and a linear mass density of
0.01kg/m. If the string is put under tension of 60N. Determine the
frequency of the third harmonic generated in the guitar string in Hz.
A. 194Hz
B. 222Hz
C. 137 Hz
D. 245 Hz
E. 173 Hz

10. Answer: A. 194Hz
We are given the information that it is third harmonic, so
m= 3
String length L = 60cm = 0.60 m (remember to convert to
meters)
μ = 0.01 kg/m
T = 60 N; so plug in the formula

11. Now we have gained
some skills in solving
basic problems. Let us do
a more in-depth problem

12. Multi-Part Problem
A 100.0cm string is vibrating with both sides clamped. The string is
kept under the tension of 25.00N. The linear mass density is 0.650g/m
a) What is the lowest frequency standing wave possible on this string?
b) What are the frequencies of the second and third harmonics
c) Can a standing wave of frequency 1.5 times the frequency
calculated in part a) be generated on this string without changing the
length / tension?
d) What is the wave speed of the string?

13. a) What is the lowest frequency standing wave possible on this string?
In order to approach this problem, the formulas needed are listed above.
For a) the question is asking for the lowest frequency. Looking at the formula,
since T ,L and µ are given and we cannot change their values, the only way to
manipulate the value of frequency is by changing the harmonic number m. So the
lowest frequency is achieved when m =1
μ = 0.650g/m = 0.00065kg/m (remember to convert to the standard unit)
L = 100.0cm = 1.000m and T = 25.00N
Plug these numbers into the first formula, we obtain the following result
Solution to a)

14. b) What are the frequencies of the second and third harmonics
For the second harmonics, only the harmonic number changes to 2, so plug
the numbers into formula 1, we get
For the third harmonics, the harmonic number changes to 3, so plug numbers
into formula 1 we get
Solution to b)

15. c) Can a standing wave of frequency 1.5 times the
frequency calculated in part a) be generated on this
string without changing the length / tension?
The answer is NO. It is tempted to just jump into the
calculation and plug numbers into the formula.
However, since we know that harmonic numbers can
only be positive integers 1,2,3… it is impossible to
get a frequency 1.5 times the fundamental frequency.
Solution to c)

16. d) What is the wave speed of the string?
In order to figure out the speed, we need frequency and wavelength. We have
already obtained the fundamental frequency in the first question as 98.1Hz.
Now we need to figure out the wavelength corresponds to the harmonic wave
when m=1.
Solution to d)

17. Note that we do not have to use the fundamental frequency from
part a). We can use the frequency of the the second and third
harmonics as long as we use the correct wavelength corresponds
to the right harmonic number
E.g. For the second harmonic, we will use m = 2 in the equation to
figure out the wavelength.
As a result, you would get the same result that v = 196m/s
Alternative Solution to d)

18. Sources
All of the equations are typed into a word
document and the screenshot is used since keynote
does not support equations
The chart is created by myself with images used
from http://physics.info/waves-standing/

19. Thank You