Question

the three waves, a, b, and c, shown in the figure below propagate on strings with equal tensions and equal mass per length. rank the waves in order of increasing frequency.

a. a<b<c
b. c=b<a
c. b<c<a
d. b<a=c
e. a=c<b

Explanation:

Step1: Recall wave - frequency relationship

The frequency $f$ and wavelength $\lambda$ of a wave are related by $v = f\lambda$, where $v$ is the wave - speed. For waves on strings with equal tensions and equal mass per length, the wave - speed $v$ is the same. So, $f=\frac{v}{\lambda}$. A shorter wavelength corresponds to a higher frequency.

Step2: Analyze the wavelengths of the waves

Wave A has the shortest wavelength among the three waves, wave B has the longest wavelength, and wave C has a wavelength in between A and B.

Step3: Rank the frequencies

Since $f=\frac{v}{\lambda}$ and $v$ is constant, the order of increasing frequency is based on the order of decreasing wavelength. So the order of increasing frequency is $B < C < A$.

Answer:

c. B < C < A