Frequency from displacement-time graph period

Steps:

• Locate one full cycle of displacement on the graph, from peak to peak or trough to trough.
• Measure the time interval for that cycle to find the period T.
• Compute frequency f using f = 1/T.
• Graph shows T ≈ 0.133 s, so f = 1/0.133 ≈ 7.5 Hz.

Why D is correct:

• Matches the definition of frequency as cycles per second, with period T = 1/f = 0.133 s directly from the graph's oscillation time.

Why the others are wrong:

• A: Implies T = 10 s, far longer than the graph's quick cycles.
• B: Implies T = 5 s, still much longer than observed.
• C: Implies T = 0.2 s, shorter than the actual period shown.

Final answer: D