Question

Two identical strings on different cellos are tuned to the 450-hertz A note. The peg holding one of the strings slips, so its tension is decreased by 2.0%. What is the beat frequency heard when the strings are then played together?

Answer 1

Given : f = 450 Hz , decrease in tension = 2 %

Solution:

As given , decrease in tension is 2 % (or 0.02)

Let , T' be the new tension and T be the old tension ,

So , (T' -T)/T = 0.02

We know that , tension is given by: T = v²

so , (v'² - v²)/v² = 0.02

And v = f

So, (²f'² - ²f²)/²f² = 0.02

(f'² -f²)f² = 0.02

f'² = 0.02*f² + f²

f' = [0.02f² +f²]^1/2

f' = [0.02(450)²+(450)²]^1/2 = 454.48 Hz

The beat frequency is given by:

f_beat = f' - f  

= 454.48 Hz - 450 Hz

= 4.48 Hz

Answer: f_beat = 4.48 Hz