Question

A $1.00\times {10}^{-20}kg$ particle is vibrating under simple harmonic motion with a period of $1.00\times {10}^{-5}s$ and with a maximum speed of $1.00\times {10}^{3}m/s$. The maximum displacement of particle from mean position is

• A. $1.59mm$
• B. $1.00mm$
• C. $10m$
• D. $3.18mm$

Answer 1

The correct option is A $1.59mm$

Given that $T=1\times {10}^{-5}s$ so $\omega =\frac{2\pi }{T}=6.28\times {10}^{5}rad/s$

The maximum speed should be $V=A\omega$

so the max displacement or $amplitude$ will be $A=\frac{V}{\omega }=\frac{{10}^3}{6.28\times {10}^5}=0.159\times {10}^{-2}meter=1.59mm$