Question Detail
Question
The 2nd overtone of a standing wave on a string of length 5 cm is 100 Hz. If the mass per unit length of the string is 2g/cm, find the tension in it.
Correct Answer
Option B is the correct answer.
Detailed Explanation
For a string fixed at both ends, the frequency of the nth harmonic is f_n = (n/2L) * sqrt(T/μ). The fundamental (n=1) is the first harmonic. The 2nd overtone is the 3rd harmonic (n=3). Length L = 5 cm = 0.05 m. μ = 2 g/cm = 0.2 kg/m. f_3 = 100 Hz. 100 = (3 / (2 * 0.05)) * sqrt(T/0.2) = (3/0.1) * sqrt(T/0.2) = 30 * sqrt(T/0.2). sqrt(T/0.2) = 100/30 = 10/3. T/0.2 = (10/3)^2 = 100/9. T = 0.2 * 100/9 = 20/9 = 2.222 N. The options are in kg/m, which is the unit for mass per unit length. The tension is in Newtons. So T = 2.222 N.
Hint
Use the formula for the frequency of a standing wave on a string: f_n = (n/2L) * sqrt(T/μ). The 2nd overtone corresponds to n=3.