Distance covered during one vibration of an oscillating body in terms of amplitude "A" is:
Correct answer: D. 4A
- A. Zero
- B. A
- C. 2A
- D. 4A
Explanation
During one complete vibration of an oscillating body in simple harmonic motion, the body travels from the equilibrium position to the maximum positive amplitude, returns to the equilibrium, continues to the maximum negative amplitude, and returns to the equilibrium position again. This results in a total distance covered of 4 times the amplitude (4A).Option A (Zero) is incorrect because the body does indeed travel a distance during the cycle. Option B (A) is incorrect because this is only the distance to the maximum displacement in one direction. Option C (2A) is incorrect as it only accounts for the distance traveled to one amplitude and back to equilibrium, which is half of the total cycle.
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Progressive waves transfer energy through oscillations, with wavelength, frequency, amplitude and wave speed related by v = fλ; sound speed depends on the medium. Superposition produces interference and stationary waves in strings and organ pipes, while simple harmonic motion describes the oscillation itself and must be distinguished from wave propagation.
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