Question

In: Physics

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers...

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 29 cm.

What is the wavelength of the sound?
Express your answer using two significant figures.

If the distance between the speakers continues to increase, at what separation will the sound intensity again be a maximum?
Express your answer using two significant figures.



Solutions

Expert Solution

Concepts and reason

The concepts used to solve this problem are weave length, the intensity of sound.

First calculate the wavelength of the sound after that calculate the separation where the intensity will maximum again.

Fundamentals

The wavelength is defined as the distance between two successive crests of wave.

The intensity of sound wave is defined as the power per square meter of the wave.

(A)

Calculate the wavelength of the sound wave.

The separation between the speakers is 19cm19{\rm{ cm}}leads maximum intensity, but separation 29cm29{\rm{ cm}}leads to zero intensity.

The waves are in phase Δx1=19cm\Delta {x_1}{\rm{ = }}19{\rm{ cm}}.

And out of the phase when Δx2=29cm\Delta {x_2} = 29{\rm{ cm}}.

Thus, the wavelength is,

Δx2Δx1=λ2\Delta {x_2} - \Delta {x_1} = \frac{\lambda }{2}

Rearrange for λ\lambda .

λ=2(Δx2Δx1)\lambda = 2\left( {\Delta {x_2} - \Delta {x_1}} \right)

Substitute 29cm29{\rm{ cm}} for Δx2\Delta {x_2} and 19cm19{\rm{ cm}} for Δx1\Delta {x_1}.

λ=2(29cm19cm)=2(10cm)=20cm\begin{array}{c}\\\lambda = 2\left( {29{\rm{ cm}} - 19{\rm{ cm}}} \right)\\\\ = 2\left( {{\rm{10 cm}}} \right)\\\\ = 20{\rm{ cm}}\\\end{array}

(B)

Calculate the separation where the sound intensity again be maximum.

As the distance between the speaker continue to increase the intensity will again be maximum when the separation between the speakers that produce a maximum has increased by one wavelength.

So,

x=Δx1+λx = \Delta {x_1} + \lambda

Substitute 19cm19{\rm{ cm}}for Δx1\Delta {x_1}, and 20cm20{\rm{ cm}}for λ\lambda .

x=(19cm)+(20cm)=39cm\begin{array}{c}\\x = \left( {19{\rm{ cm}}} \right) + \left( {20{\rm{ cm}}} \right)\\\\ = 39{\rm{ cm}}\\\end{array}

Ans: Part A

The wavelength of the sound is20cm20{\rm{ cm}}.

Part B

The separation where the sound intensity again be maximum is 39cm39{\rm{ cm}}.


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