Standing Waves on Strings (14-13)

1. Physics 101 Learning Object (LO6)
By: James Kurniawan

2.  As this string gets plucked, waves travel along it,
reflecting from the 2 fixed ends.
 Travelling waves moving in opposite directions,
resulting in STANDING WAVES
Image from : http://theory.uwinnipeg.ca/physics/bohr/img62.gif

3. Wavelength?
 Because string is clamped at both ends: Amplitude
must be 0 at x=0 and x=L -
 and
 Rearranging and simplifying the equation above :
 These standing waves = NORMAL MODES of
vibration of string
*longest wavelength possible is 2L.
*m = essentially ANTINODES
Equations are from textbook: 14-48 to 14-50
*m can only be a positive,
nonzero INTEGER

4. Frequency?
 From the normal modes :
 Lowest frequency = longest wavelength λ1 (2L)
 FUNDAMENTAL FREQUENCY or FIRST
HARMONIC
*note : v=√(T/µ) is from
Equation (14-5), Section 14-
4. Proportional to square
root of tension, Inversely
proportional to length and
square root of linear mass
density
Higher frequencies = higher values of m* from
the fundamental frequency
*can only be POSTIVE, NONZERO INTEGERS!
Equations and image are from textbook : 14-51 and 14-52

5. Harmonics
 Allowed frequencies (where f=mf , and m = integers) =
HARMONICS or RESONANT FREQUENCIES
 A string can vibrate in a single normal mode or in
several modes
Equation is from textbook : 14-54

6. Question
 The string on the bottom is 5.0 meters long and is
vibrating as the fourth harmonic. Its linear mass
density is 4.25 x 10^-3 kg/m. The tension in the string
is kept at 60.0N.
 Find its wavelength and frequency.
Picture from :
http://www.physicsclassroom.com/class/waves/Lesson-
4/Mathematics-of-Standing-Waves

7. Answer (Part 1)
1) Wavelength:
Using Equation (14-50) : λm = 2L/m
- m = 4, L = 5.0 m
 λ4 = 2(5.0)/(4) = 2.5 m

8. Answer (Part 2)
2) Frequency :
Using Equation (14-51) : fm = m/2L * √(T/µ)
- m = 4, L = 5.0 m, T = 60.0 N,
µ = 4.25 x 10^-3 kg/m
 f4 = (4)/2(5.0) * √(60.0/4.25 x 10^-3)
= 47.5 Hz