A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the fundamental frequency?
A) 316 Hz
B) 632 Hz
C) 158 Hz
D) 215 Hz
E) 79 Hz

Question

Quizplus · Accepted Answer

The fundamental frequency of a string fixed at both ends is given by:
f = (1/2L) * sqrt(Tension/linear density)

Where L is the length of the string (in this case 200 cm), Tension is the tension applied to the string (600 N), and linear density is the density of the string (0.015 g/cm).

Substituting the values given, we get:
f = (1/2*200cm) * sqrt(600N/(0.015 g/cm)) 
f = 158 Hz (approx)

Therefore, the answer is option C, with a fundamental frequency of 158 Hz.

A String Is Stretched and Fixed at Both Ends, 200