A guitar string of length 0.8m and mass m=8g is suspended at two fixed ends unde

Question

A guitar string of length 0.8m and mass m=8g is suspended at two fixed ends under a tension force of 130N. A person plucks the string which generates a standing wave of the lowest harmonic on the string.

(a) What is the wavelength of the standing wave?

(b) What is the frequency of the generated tone (do not use the speed of sound for this calculation!)?

(c) If the generated sound wave carries a total power of 3 x 10-9 W, and if it propagates spherically, what is the intensity level (in dB) at a distance of 7m from the source?

PLEASE SHOW WORK!

Explanation / Answer

the string is vibrating inthe fundamental mode at its lowest frequency

linear density = u = m/L = 0.008/0.8 = 0.01 kg /m

speed = v = sqrt(T/u) = sqrt(130/0.01) = 114.0175 m/s

part (a) wavelength = 2L = 2*0.8 = 1.6 m

part (b) frequency fo = v/wavelength = 71.26 Hz

part(c)

intensity = I = power/ares

area = 4*pi*r^2 = 4*3.14*7*7 = 615.44 m^2

I = 3*10^-9/(615.44)

I = 4.8746*10^-12 W/m^2

in decibels intensity = 10*log(I/Io)

Io = threshold of hearing = 10^-12

I = 10*log4.8746

I = 6.87 dB <----answer

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