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.

wave 1

wave 2

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: Analyze Amplitude

Amplitude is the maximum vertical displacement from equilibrium. For Wave 1, the peak is at \( y = 1.5 \) (approx), for Wave 2, peak is at \( y = 1 \) (approx). So amplitude is greater for Wave 1.

Step2: Analyze Wavelength

Wavelength is the distance between two consecutive peaks (or troughs). Wave 1: from \( x = 0 \) to \( x = 10 \) (one wavelength? Wait, no, looking at the graphs: Wave 1 has a peak at ~2, trough at ~7, peak at ~12, so distance between peaks (0 to 10? Wait, the x-axis: Wave 1, from 0 to 10, it completes one full wave? Wait, no, let's check the positions. Wave 1: at x=0, it's 0; x=5, 0; x=10, 0; x=15, 0; x=20, 0. Wait, the number of cycles: Wave 1 has 2 cycles (from 0-10, 10-20? Wait, no, the first peak is around x=2, then trough at x=7, peak at x=12, trough at x=17. So wavelength is distance between two peaks: 12 - 2 = 10? Wait, Wave 2: peaks at x=3, x=13, so distance is 10? Wait, no, maybe I'm miscalculating. Wait, the x-axis for both goes to 20. Wave 1: how many cycles? From 0 to 20, it has 2 cycles (since from 0-10, one cycle, 10-20 another? Wait, no, the first peak is at ~2, then trough at ~7, peak at ~12, trough at ~17. So two cycles in 20 units? No, 0-10: peak at 2, trough at 7, back to 0 at 10. Then 10-20: peak at 12, trough at 17, back to 0 at 20. So each cycle is 10 units (wavelength = 10 ft). Wave 2: peaks at ~3, ~13, so wavelength is 10 ft? Wait, no, Wave 2: from 0-5, up to peak, down to 0 at 5; then down to trough at 7.5, up to 0 at 10; then up to peak at 12.5, down to 0 at 15; then down to trough at 17.5, up to 0 at 20. So Wave 2 has 2 cycles in 20 units? Wait, no, 0-10: one cycle (up, down, up? No, 0-5: up, 5-10: down, 10-15: up, 15-20: down. So two cycles in 20 units? Wait, no, wavelength is distance between two consecutive peaks (or two consecutive troughs). For Wave 1: peaks at x=2 and x=12, so distance is 10. For Wave 2: peaks at x=3 and x=13, distance is 10. Wait, maybe I'm wrong. Wait, the first graph (Wave 1) has a larger amplitude (higher peaks) than Wave 2. Now wavelength: let's count the number of cycles in the same distance. Both graphs go to x=20. Wave 1: how many full cycles? From 0 to 20, it has 2 cycles (since from 0-10, it goes up, down, back to 0; 10-20 same). Wave 2: from 0-20, it has 2 cycles? Wait, no, Wave 2: at x=0, 0; x=5, 0; x=10, 0; x=15, 0; x=20, 0. Wait, the number of peaks: Wave 1 has 2 peaks (at ~2 and ~12), Wave 2 has 2 peaks (at ~3 and ~13)? No, wait Wave 2's graph: the first peak is lower, then trough, then peak, etc. Wait, maybe the wavelength is the same? Wait, no, let's check the distance between two consecutive zero crossings (where it crosses the x-axis going up). For Wave 1: crosses at 0, 5, 10, 15, 20. So the distance between 0 and 10 is 10, which is two half-waves? No, wavelength is the distance between two identical points, like peak to peak. For Wave 1, peak at x=2, next peak at x=12: distance 10. For Wave 2, peak at x=3, next peak at x=13: distance 10. So wavelength is equal? Wait, no, maybe I'm miscalculating. Wait, the problem says "assuming the waves are traveling at the same speed". Now frequency: \( f = \frac{v}{\lambda} \), where \( v \) is speed, \( \lambda \) is wavelength. If \( v \) is same, and \( \lambda \) is same, then frequency same? Wait, no, wait let's count the number of cycles. Wave 1: from 0 to 20, how many cycles? Let's see the graph: Wave 1 has two full cycles (up, down, up, down) in 20 ft? Wait, no, the first wave (Wave 1) has a larger amplitude (the height from equilibrium is more). Wave 2 has smaller amplitude. Now wavelength: let's see the distance between two consecutive peaks. Wave 1: peak at x=2, then next peak at x=12: so wavelength \( \lambda_1 = 10 \) ft. Wave 2: peak at x=3, next peak at x=13: \( \lambda_2 = 10 \) ft. So wavelength is equal. Now frequency: \( f = \frac{v}{\lambda} \). Since \( v \) is same and \( \lambda \) is same, frequency is equal? Wait, no, wait maybe I made a mistake. Wait, let's count the number of cycles in the graph. Wave 1: from 0 to 20, how many times does it go up and down? Let's see: starts at 0, goes up to peak, down to trough, up to 0 at x=5? No, x=5 is 0. Then down to trough, up to peak, down to 0 at x=10? No, x=10 is 0. Then down to trough, up to peak, down to 0 at x=15? No, x=15 is 0. Then down to trough, up to 0 at x=20. Wait, maybe each wave has two cycles in 20 ft. So wavelength is 10 ft for both. Then amplitude: Wave 1's peak is higher (closer to 2) than Wave 2's (closer to 1). So amplitude is greater for Wave 1. Wavelength: same (since both have two cycles in 20 ft, so wavelength 10 ft). Frequency: \( f = \frac{v}{\lambda} \), so same frequency.

Step3: Analyze Frequency

Since \( v = f\lambda \), and \( v \) is same, \( \lambda \) is same, so \( f \) is same.

Answer:

• The amplitude is: greater for Wave 1
• The wavelength is: equal for both waves
• The frequency is: equal for both waves

(To mark the circles: For "The amplitude is", mark the circle under "greater for Wave 1". For "The wavelength is", mark the circle under "equal for both waves". For "The frequency is", mark the circle under "equal for both waves".)