Question

the graphs below model two transverse waves that maya created with a battle rope. in each graph, the horizontal axis shows the horizontal position of particles of the rope, measured from where maya stood. the vertical axis shows the vertical displacement of the rope, measured from the equilibrium position, at a specific moment in time.

graphs of wave 1 and wave 2 are shown here

assuming the waves are traveling at the same speed, compare the waves.

	greater for wave 1	greater for wave 2	equal for both waves
the wavelength is	∘	∘	∘
the frequency is	∘	∘	∘

Explanation:

Step1: Determine the amplitude

Amplitude is the maximum displacement from the equilibrium position. For Wave 1, the amplitude is \(1\) ft (from \(y = 0\) to \(y=1\)). For Wave 2, the amplitude is also \(1\) ft (from \(y = 0\) to \(y = 1\)). So the amplitude is equal for both waves.

Step2: Determine the wavelength

Wavelength is the distance between two consecutive crests (or troughs). For Wave 1, the wavelength \(\lambda_1=10\) ft (from \(x = 0\) to \(x = 10\) for one full cycle). For Wave 2, the wavelength \(\lambda_2 = 5\) ft (from \(x=0\) to \(x = 5\) for half - cycle, so full - cycle is \(5\) ft). So the wavelength is greater for Wave 1.

Step3: Use the wave speed formula \(v=\lambda f\)

Since \(v_1 = v_2\) (given \(v\) is the same), and \(v=\lambda f\), \(f=\frac{v}{\lambda}\). If \(\lambda_1>\lambda_2\), then \(f_1

Answer:

The amplitude is equal for both waves; The wavelength is greater for Wave 1; The frequency is greater for Wave 2.