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BIO For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about 6.0×10-5Pa. Calculate the (a) intensity; (b) sound intensity level; (c) displacement amplitude of this sound wave at 20°C.

Short Answer

Expert verified

A) I=4.36×10-12W/m2 b) β=6.4dB c) A=5.8×10-11m

Step by step solution

01

Concept of the intensity of the sound wave in terms of the pressure amplitude

Formula isI=vp2max2D isthe relation that describes the intensity of the sound wave in terms of the pressure amplitude

02

Calculate the intensity of he sounds

Substitute the values in the equationI=vp2max2Dwe get,

I=344×6×10-522×1.42×105=4.36×10-12W/m2

The relation that describes the intensity level of a sound wave isβ=10logll0 .Substitute the values in the above equation we get,

β=10log4.36×10-12W/m210-12W/m2=6.4dB

Therefore, the intensity of the sound is 6.4dB

03

Step 3 The relation between the displacement amplitude and the pressure amplitude

The equation is A=PmaxkB

We know that the Bulk modulus for the air is B=1.42×105Paand the pressure amplitude is 6×10-5Pabut in order to make use of equation (3) we need to determine k, where k can be represented as follows

k=2πλλ=vf=344400=0.86mk=2π0.86m=7.3rad/mA=6×10-5Pa7.3rad/m×1.42×105Pa=5.8×10-11m

Therefore, the pressure amplitude is5.8×10-11m

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