Question

Rank in order, from largest to smallest, the wavelengths ${\lambda }_a,{\lambda }_{b}and{\lambda }_c$for sound waves having frequencies $f_a=100Hz,f_b=1000Hz,and{f}_c=10,000Hz$. Explain.

Short answer

The largest wavelength is ${\lambda }_a$and the smallest wavelength is ${\lambda }_c$.

${\lambda }_a>{\lambda }_b>{\lambda }_c$

Step-by-step solution

Step 1. Write the given information.

The wavelengths of the sound wave are given by ${\lambda }_a,{\lambda }_{b}and{\lambda }_c$
The frequency of waves is role="math" localid="1648816875381" $f_a=100Hz,f_b=1000Hz,and{f}_c=10,000Hz$
respectively.

The speed of the wave is given by the formula $\lambda f=c$
Here, c is the speed of light (3x10⁸m/s²) which is constant.

Step 2. To determine the wavelengths of the given wave

When the frequency of the wave is $f_a=100Hz$

The wavelength is determined by ${\lambda }_a{f}_a=c$
$\Rightarrow {\lambda }_a=\frac{c}{f_a}=\frac{3\times {10}^8}{100}=3\times {10}^{6}m$

Similarly, for frequency $f_b=1000Hz$

The wavelength is given as;

${\lambda }_b=\frac{c}{f_b}=\frac{3\times {10}^8}{1000}=3\times {10}^{5}m$

And, for frequency $f_c=10,000Hz$

The wavelength is given by;

${\lambda }_c=\frac{c}{f_c}=\frac{3\times {10}^8}{10,000}=3\times {10}^{4}m$

Thus, it can be observed that ${\lambda }_a>{\lambda }_b>{\lambda }_c$
The largest wavelength is ${\lambda }_a$and the smallest wavelength islocalid="1648817329905" ${\lambda }_c$.