Question
A particle executes simple harmonic motion. The amplitude of vibration of particle is $2 \mathrm{~cm}$. The displacement of particle in one time period is :(a) $1 \mathrm{~cm}$(b) $2 \mathrm{~cm}$(c) $4 \mathrm{~cm}$(d) zero
Step 1
The particle starts from a point and moves in one direction to a maximum distance (amplitude) and then moves back to the starting point. This is half of the time period. In the next half, the particle moves in the opposite direction to the same maximum distance Show more…
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Round 2
A particle is vibrating in a simple harmonic motion with and amplitude of $4 \mathrm{~cm}$. At what displacement from the equilibrium position is its energy half potential and half kinetic? (a) $1 \mathrm{~cm}$ $\begin{array}{ll}\text { (b) } \sqrt{2} \mathrm{~cm} & \text { (c) } 3 \mathrm{~cm}\end{array}$ (d) $2 \sqrt{2} \mathrm{~cm}$
Round 1
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