If the total energy of a harmonic oscillator is reduced to 1/3 of its original value, what is the change in the amplitude of the oscillations?

Question

BlackTom AI · Accepted Answer

Begin by recalling a key relationship for a simple harmonic oscillator: the total energy E is proportional to the square of the amplitude, E ∝ A^2, since E = (1/2) k A^2 for a spring with spring constant k.

If the total energy is reduced to 1/3 of its original value, the new energy E' = (1/3) E.

Because E ∝ A^2, the new amplitude A' must satisfy (A')^2 ∝ E' = (1/3) E ∝ (1/3) A^2. Taking square roots gives A' = A / √3.

Therefore, the amplitude changes by a factor of 1/√3 relative to the original amplitude. In decimal form, 1/√3 ≈ 0.577, so the amplitude is about 57.7% of its original value.

Now looking at the provided answer options: the two given possibilities appear to be 1 and 3 (based on the formatting "1
3"), which do not correspond to the required factor of 1/√3. This indicates that neither option matches the correct mathematical result. The mismatch could arise from a misprint or an incorrect framing of the choices.

Since the question asks for the change in amplitude, the appropriate description is that A' = A / √3, not simply a change by a factor of 1 or 3. If one were to express the change in terms of a ratio, the correct ratio is 1/√3, which is less than 1, indicating a reduction in amplitude.

If you were expected to pick from two options, you would need options that reflect the 1/√3 factor or its numerical approximation, not 1 or 3. The missing or mismatched options prevent selecting a correct choice from the presented set.

SPHY001 Practice Quiz - Unit 16 | LA056

If the total energy of a harmonic oscillator is reduced to 1/3 of its original value, what is the change in the amplitude of the oscillations?

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