# For sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2 x 10^( 8) m...

## Question:

For sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2{eq}\times {/eq}10{eq}^{-8} {/eq} m produces a pressure amplitude of 3.0{eq}\times {/eq}10{eq}^{-2} {/eq} Pa. Use Vsound = 344 m/s.

a) For 1000 Hz waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa?

b) For what wavelength will waves with a displacement amplitude of 1.2{eq}\times {/eq}10{eq}^{-8} {/eq} m produce a pressure amplitude of 1.5{eq}\times {/eq}10{eq}^{-3} {/eq} Pa?

c) For what frequency will waves with a displacement amplitude of 1.2{eq}\times {/eq}10{eq}^{-8} {/eq} m produce a pressure amplitude of 1.5{eq}\times {/eq}10{eq}^{-3} {/eq} Pa?

## Sound:

Sound is defined as a mechanical wave that travels in different mediums like solid liquid and gas and has different speeds. Sound travels faster in the solid and slower in the liquid, and the slowest in the gases. Its frequency, amplitude and wavelength can characterize sound.

## Answer and Explanation: 1

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Given data,

Displacement amplitude, {eq}{A_1} = 1.2 \times {10^{ - 2}}\,{\rm{m}} {/eq}

Maximum pressure amplitude, {eq}{P_{1\,\max }} = 3 \times...

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