Question

Tuning an Instrument. A musician tunes the C-string of her instrumeut to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?

Answer 1

(a) The tension the musician must stretch it is 147.82 N.

(b) The percent increase in tension is needed to increase the frequency is 26%.

Tension in the string

v = √T/μ

where;

• v is speed of the wave
• T is tension
• μ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/m

v = Fλ

in fundamental mode, v = F(2L)

v = 2FL

v = 2 x 65.4 x 0.6 = 78.48 m/s

v = √T/μ

v² = T/μ

T = μv²

T = 0.024 x (78.48)²

T = 147.82 N

When the frequency is 73.4 Hz;

v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s

T = μv²

T = (0.02)(88.08)²

T = 186.19 N

Increase in the tension

= (186.19 - 147.82)/(147.82)

= 0.26

= 0.26 x 100%

= 26 %

Thus, the tension the musician must stretch it is 147.82 N.

The percent increase in tension is needed to increase the frequency is 26%.

Learn more about tension here: brainly.com/question/24994188

#SPJ1