Intensity proportional to amplitude squared Steps:

• Intensity $I \propto A^2$ for sound waves, where $A$ is displacement amplitude.
• Halving intensity means new amplitude $A' = A / \sqrt{2}$, reducing peak displacement by $\sqrt{2}$ factor.
• Constant frequency keeps oscillation period unchanged, so waveform shape and time scale remain the same.
• New graph shows same frequency but smaller amplitude.

Why D is correct:

• D depicts reduced amplitude by $\sqrt{2}$ with unchanged frequency, matching $I \propto A^2$.

Why the others are wrong:

• A halves amplitude linearly, ignoring square root relation.
• B alters frequency, contradicting constant frequency.
• C shows no change in amplitude, ignoring intensity halving.

Final answer: D