A string that is 0.15 m long and fixed at both ends is vibrating in its n = 5 harmonic.The sound from this string excites a pipe that is $0.82 \mathrm {~m}$ long and open at both ends into its second overtone resonance.What is the distance between a node and an adjacent antinode,in the string? The speed of sound in air is 345 m/s.
A)15 mm
B)7.5 mm
C)23 mm
D)30 mm
E)150 mm

Question

Quizplus · Accepted Answer

For the string vibrating in its n = 5 harmonic, the wavelength $\lambda$ can be found using the relationship that the length of the string $L$ is equal to $n\lambda/2$, where $n$ is the harmonic number. Given $L = 0.15$ m and $n = 5$, we have $0.15 = 5\lambda/2$, solving for $\lambda$ gives $\lambda = 0.06$ m. The distance between a node and an adjacent antinode is $\lambda/4$, which is $0.06/4 = 0.015$ m or 15 mm.

A String That Is 0 0.82 m0.82 \mathrm {~m}0.82 m Long and Open at Both Ends into Its Second Overtone

A String That Is 0 $0.82 \mathrm {~m}$ Long and Open at Both Ends into Its Second Overtone