A 25.0-g string is stretched with a tension of 43.0 N between two fixed points 12.0 m apart. What is the frequency of the second harmonic?
A) 6.00 Hz
B) 12.0 Hz
C) 18.0 Hz
D) 24.0 Hz
E) 36.0 Hz

Question

Quizplus · Accepted Answer

The frequency of a string fixed at both ends can be calculated using the formula:

f = (n/2L) * sqrt(T/m)

where:
- n is the harmonic number
- L is the length of the string
- T is the tension in the string
- m is the mass per unit length of the string

For the second harmonic (n=2), we have:

f = (2/2L) * sqrt(T/m)

Substituting the given values, we get:

f = (1/L) * sqrt(T/m)
  = (1/12.0 m) * sqrt(43.0 N / (25.0 g / 12.0 m))
  = 12.0 Hz

Therefore, the answer is B) 12.0 Hz.

A 250-G String Is Stretched with a Tension of 43