The frequency of a Guitar string, f, multiplying by the wave length on the strin

Question

The frequency of a Guitar string, f, multiplying by the wave length on the string equal the wave velocity of the standing wave on the string. The wave velocity in the string equals squared root of Tension/(mass/total length). Now there are two strings of same length made same frequency and same amplitude, but the ratio mass, to mass, is 2, tension of string 1 is 8 times of that on string it. A) Drive the wave velocity in string I. Formula: ___ B) Drive the velocity on string II, Formula ____ C) What is the ration of V_1 to V_2. D) By using V = f x lambda, where lambda is the wavelength, find the ratio of two wave lengths, Formula:____ E) Draw the two transverse wave in the following section of guitar strings. string {} |.....| for longer wavelength with half wavelength string {} |.....| for the short wavelength The intensity of a sound or light is the power divided by the area that received the power. The of watts/meter^2 is for the intensity. Since the range of power reached may be several ten's power magnitude then we need to take log to make smaller number for the intensity, so called decibel 10 times log (I/I_0) where I_0 is the minimum intensity we can hear, that is 1 times 10^-12 watt/m^2. Now use a normal conservation intensity of 1 times 10^-6 watt/m^2 a) Find the ratio of I/I_0 answer: ___ b) Take the log of the intensity compared with minimum intensity.

Explanation / Answer

Q.8. For heat engine

Qin = 11410 J

W = 2500 J

(A) We know that, W = Qin - Qout,

hence Qout = Qin - W = 11410 - 2500 = 8910 J

(B) Efficiency of engine = W / Qin = 2500 / 11410 = 21.9106 %

(C) Efficiency of engine = 1 - (T_low / T_high) = 1 - (200/2500) = 8 %

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