A 69 cm electric-guitar string (linear density = 0.0515 g/cm) is under 186.5 N t

Question

A 69 cm electric-guitar string (linear density = 0.0515 g/cm) is under 186.5 N tension and is out of tune for A(440 Hz). The ambient temperature is 23C. A 440 Hz tuning fork is used to tune guitar to 440 Hz. 12 beats are heard over a 3 s interval, and the string is tightened until no beats are heard. The frequency of the un-tuned string was The change in string tension to tune the string to 440 Hz is The wavelength of the 4th harmonic played on the tuned string is The frequency of the sound produced (in air) is The wavelength of the sound wave is m. The sound level 3m from the loudspeaker is 60 dB. The sound level at the back of the auditorium 60 m from the loudspeaker is

Explanation / Answer

iven data

mue = 0.0515 g/cm = 0.00515 kg/m

T1 = 186.5 N

T = 23 C

beat frequency, f_beat = 12/3 = 4 hz

A1) The frequency untunedtring was = 440 - 4 = 436 hz

A2) v = lamda*f

lamda = v/f

lamda1 = lamda2

v1/f1 = v2/f2

v1/v2 = f1/f2

sqrt(T1/T2) = f1/f2

T1/T2 = (f1/f2)^2

==> T2 = T1*(f2/f1)^2

= 186.5*(440/436)^2

= 190 N

B1) wavelength of 4th harmonic = 2*L/4

= 2*0.69/4

= 0.345 m

C1) 440 hz

C2) speed of sound in air, v = vo + T*0.6

= 331 + 25*0.6

= 346 m/s

now use relation, v = lamda*f

==> lamda = v/f

= 346/440

= 0.786 m

D1)

we know, Intelsity = power/Area

I = P/(4*pi*d^2)

so, I1/I2 = d2^2/d1^2

I2 = I1*(d1/d2)^2

= I1*(3/60)^2

= I1*0.0025

at 3m,

sound level
beta1 = 10*log(I1/Io)

60 dB = 10*log(I1/Io)

at 60m,

sound level
beta2 = 10*log(I2/Io)

= 10*log(I1*0.0025/Io)

= 10*log(I1/Io) + 10*log(0.0025)

= 60 + 10*log(0.0025)

= 33.98 dB

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