Standing Waves On Strings: Harmonic and Frequency Problem

[K_Physics]

Homework Statement

String A is stretched between two clamps separated by distance L. String B, with the same linear density and under the same tension as string A. String B is stretched between two clamps separated by distance 4L. Consider the first eight harmonics of string B. For which of these eight harmonics of B (if any) does the frequency match the frequency of (a) A’s first harmonic, (b) A’s second harmonic, and (c) A’s third harmonic?

Not sure if I correctly solved the problem (hopefully I did =D). Just need someone to check over my work =D. Thanks!

Homework Equations

ν = √(T/μ)

ν = ƒλ

L = nλ/2

ν: Velocity
T: Tension
μ: Linear Density
ƒ: Frequency
λ: Wavelength
L: Length
n: nth Harmonic[/B]

The Attempt at a Solution

[/B]
Since the tension and linear density on both strings are equal, the velocity is also equal.

Next I solved for the frequency of system B:

ν = ƒλ
ƒ = ν/λ → 1

L = nλ/2, since L = 4L

λ = 8L/n → 2

Subbing 2 → 1

ƒ = νn/8L

--------------------------------------------------------------------------------
First Harmonic of String A:

L= λ/2 ⇒ λ = 2L
ƒ=ν/λ ⇒ ƒa1 = ν/2L

ƒa1 = ƒb

ν/2L = νn/8L
n = 4 (between 1-8)

Second Harmonic of String A:

L= λ
ƒ=ν/λ ⇒ ƒa2 = ν/L

ƒa2 = ƒb

ν/L = νn/8L
n = 8 (between 1-8)

Third Harmonic of String A:

L= 3λ/2 ⇒ λ = 2L/3
ƒ=ν/λ ⇒ ƒa3 = 3ν/2L

ƒa3 = ƒb

3ν/2L = νn/8L
n = 12 (not between 1-8)

 

[TSny]

Hello. Welcome to PF!

Your work looks correct.

A similar approach is to note that the harmonic frequencies of A are $f_A = \frac{n_ {_A} \ v}{2L}$ while those of B are $f_B = \frac{n_{_B} \ v}{8L}$.

Show that $f_A =f_B$ implies $n_{_B} = 4n_{_A}$. Then let $n_{_A}= 1$ for the first harmonic of A, etc.

 

[K_Physics]

Thanks for the help! I will try the question both ways =D.