# a. What are the three longest wavelengths for standing waves

a. What are the three longest wavelengths for standing waves on a 240-cm-long string that is fixed at both ends?
b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Heather Fulton
Step 1
a) The three longest wavelengths for standing waves will therefore correspond to m 1, 2, and 3. Thus
${\lambda }_{1}=\frac{2\left(2.40m\right)}{1}=4.80m$
${\lambda }_{2}=\frac{2\left(2.40m\right)}{2}=2.40m$
${\lambda }_{3}=\frac{2\left(2.40m\right)}{3}=1.60m$
Step 2
b) Because the wave speed on the string is unchanget from one m value to the other,
${f}_{2}{\lambda }_{2}={f}_{3}{\lambda }_{3}⇒{f}_{3}=\frac{{f}_{2}{\lambda }_{2}}{{\lambda }_{3}}=\frac{\left(50.0Hz\right)\left(2.40m\right)}{1.60}=75.0Hz$

xandir307dc
Step 1
a) Knowing the formula for the wavelength of a standing wave with fixed nodes at the ends,
${\lambda }_{m}=\frac{2L}{m}$
we can easily get
${\lambda }_{1}=\frac{2×240}{1}=480m$
${\lambda }_{2}=\frac{2×240}{2}=240cm$
${\lambda }_{3}=\frac{2×240}{3}=160cm$
b) We know that the frequency of an m-mode number standing wave is
${f}_{m}=m{f}_{1}$
Therefore, we have
$\frac{{f}_{3}}{{f}_{2}}=\frac{3}{2}⇒{f}_{3}=\frac{3{f}_{2}}{2}=75Hz$