Question

At a certain temperature and pressure, the speed of sound in helium is 988 meters per second, and the speed of sound in air is 334 meters per second. A sound with a wavelength of 4.93 meters travels in helium and passes into air. What are the wavelength and frequency of the sound after it enters the air? Include units in your answers. Answer must be in 3 significant digits.

Answer 1

We are given that the speed of sound in helium is 988 m/s. We are also given that a sound has a wavelength of 4.93 meters. We can determine the frequency of this sound using the following equation:

`v_(He)=f\lambda_(He)`

Where:

`\begin{gathered} v_(He)=\text{ sp}eed\text{ of sound in Helium} \\ f=frequency \\ \lambda_(He)=\text{ wavelength in Helium} \end{gathered}`

Now, we can solve for the frequency by dividing both sides by the wavelength:

`(v_(He))/(\lambda_(He))=f`

Now, we plug in the values:

`(988(m)/(s))/(4.93m)=f`

Solving the operations:

`200.406Hz^{}=f`

Now, the frequency of sound depends on the source and therefore remains constant independent of the medium, therefore, the frequency of the sound as it passes from Helium to air is the same, therefore the frequency in the air is:

`f=200.406Hz`

Now, we use the formula for air:

`v_{\text{air}}=f\lambda_(air)`

Now, we solve for the wavelength by dividing both sides by the frequency:

`\frac{v_{\text{air}}}{f}=\lambda_(air)`

Now, we plug in the values:

`(334(m)/(s))/(200.406Hz)=\lambda_(air)`

Solving the operations:

`1.667m=\lambda_(air)`

Therefore, the wavelength in air is 1.667 meters.