2. WHAT IS A STANDING WAVE?
• When two harmonic waves of equal amplitude,
wavelength, frequency are moving in opposite direction
3. WHAT IS A STANDING WAVE?
Node = no motion at 0 amplitude
Antinode = maximum possible amplitude of 2
The node and antinode are λ/4 far apart.
λ/4
4. KEY EQUATIONS
• D(x,t) = 2Asin(kx)cos(ωt)
• Location of node
• When sin(2πx/λ) = 0, x = nπ
• Location of antinode
• When sin(2πx/λ) = ±1, x = (n+1/2) λ/2
5. KEY EQUATIONS
• Standing Wave on a string
• Fixed end of string -> length (L) = x
• λ = 2L/n where n = number of antinode in integers
• f = v/λ = nv/2L = n/2L * √T/μ
• fn = nf1 where f1 = fundamental frequency of a string
6. • Application of standing wave on instruments
• Both end closed/opened
• λ = 2L/n
• f = nv/2L = n/2L * √T/μ
• One end open and another one closed
• λ = 4L/n
• f = nv/4L = n/4L * √T/μ
KEY EQUATIONS
7. QUESTION 1
• A guitar string oscillates in a standing wave pattern after a
person plays with it.
• D(x,t) = 1.5sin(0.6x)cos(20πt)
• 1. What is the wavelength and the frequency?
• 2. What is the distance between two consecutive nodes?
8. SOLUTION
1. The wavelength (λ)
• D(x,t) = 1.5sin(6x)cos(30πt)
• D(x,t) = 2Asin(kx)cos(ωt)
• k = 2π/λ, λ= 2π/k = 2π/6 = 1.05 m
The frequency (f)
ω = 2πf , f = ω/2π = 30π/2π = 15 Hz
9. 2. The distance between two consecutive nodes
• λ = 2L/n where n = 1
• When there is one antinode, there are two nodes at both
ends
• The distance between two nodes is L.
• L = λ/2 = 1.05 m / 2 = 0.525 m
SOLUTION
11. QUESTION 2
• The guitar string has the length of 1.5m long fixed at both ends.
• It has a frequency of 300 Hz. The tension is increased by 10%.
• What is the new fundamental frequency of the string.
12. SOLUTION
• f = nv/2L = n/2L * √T/μ
• When the tension is increased by 10%, the T will increase to
1.1T
• f = n/2L * √T/μ = ½(1.5m)*√T/μ = 0.75√T/μ = 300 Hz
• f = 0.75√1.1*T/μ = 0.75√T/μ * √1.1 = 300 * √1.1 = 315 Hz
