look at the table you completed.\nwhich wave transmits more energy per second?

Explanation:

Step1: Recall the formula for wave energy

The energy transmitted by a wave per second (power) is given by (P = \frac{1}{2}\mu A^{2}\omega^{2}v), where (\mu) is the linear mass density (assumed same for both waves), (A) is the amplitude, (\omega = 2\pi f) ( (f) is frequency), and (v) is the wave - speed (assumed same for both waves). Since (P\propto A^{2}f^{2}) (because (\mu) and (v) are constant).

Step2: Analyze the amplitude

The amplitude (A) is the maximum displacement from the equilibrium position. For Wave 1 and Wave 2, from the graphs, the amplitude of Wave 1 and Wave 2 is the same.

Step3: Analyze the frequency

The frequency (f=\frac{v}{\lambda}) (where (\lambda) is the wavelength). From the position - displacement graph, the wavelength (\lambda) (distance between two consecutive crests or troughs) of Wave 1 is larger than that of Wave 2. Since (v) is constant (assuming same medium), (f=\frac{v}{\lambda}), so (f_2>f_1) (because (\lambda_1 > \lambda_2)).

Since (P\propto A^{2}f^{2}) and (A_1 = A_2) and (f_2>f_1), Wave 2 transmits more energy per second.

Answer:

Wave 2 transmits more energy per second.