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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Simple Harmonic Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Simple Harmonic Motion

During a simple harmonic motion, which of the following quantities...

During a simple harmonic motion, which of the following quantities will always be negative?

Options

  1. AF⃗⋅a⃗\vec{F} \cdot \vec{a}
  2. Bv⃗⋅r⃗\vec{v} \cdot \vec{r}
  3. CF⃗⋅r⃗\vec{F} \cdot \vec{r}
  4. Da⃗⋅r⃗\vec{a} \cdot \vec{r}

Correct answer

D. a⃗⋅r⃗\vec{a} \cdot \vec{r}

Step-by-step solution

In simple harmonic motion, the restoring force F⃗\vec{F} and acceleration a⃗\vec{a} are always directed towards the mean position, while the displacement r⃗\vec{r} is directed away from the mean position.

Mathematically, F⃗=−kr⃗\vec{F} = -k\vec{r} and a⃗=−ω2r⃗\vec{a} = -\omega^2\vec{r}, where kk and ω2\omega^2 are positive constants.

Evaluating the given dot products:

F⃗⋅a⃗=(−kr⃗)⋅(−ω2r⃗)=kω2r2≥0\vec{F} \cdot \vec{a} = (-k\vec{r}) \cdot (-\omega^2\vec{r}) = k\omega^2 r^2 \ge 0

v⃗⋅r⃗\vec{v} \cdot \vec{r} can be positive (when the particle is moving away from the mean position) or negative (when moving towards the mean position).

F⃗⋅r⃗=(−kr⃗)⋅r⃗=−kr2≤0\vec{F} \cdot \vec{r} = (-k\vec{r}) \cdot \vec{r} = -kr^2 \le 0

a⃗⋅r⃗=(−ω2r⃗)⋅r⃗=−ω2r2≤0\vec{a} \cdot \vec{r} = (-\omega^2\vec{r}) \cdot \vec{r} = -\omega^2 r^2 \le 0

Thus, F⃗⋅r⃗\vec{F} \cdot \vec{r} and a⃗⋅r⃗\vec{a} \cdot \vec{r} will always be negative for any non-zero displacement.
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