Question

The figure shows a graph between velocity and displacement (from mean position) of a particle performing SHM:

• A. The time period of the particle is 1.57s
• B. The maximum acceleration will be $40cm/s^2$
• C. The velocity of a particle is $2\sqrt{2}1cm/s$ when it is at a distance of 1 cm from the mean position
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A The time period of the particle is 1.57s
B The maximum acceleration will be $40cm/s^2$
C The velocity of a particle is $2\sqrt{2}1cm/s$ when it is at a distance of 1 cm from the mean position

by graph we can see that $V_{max}=10cm/s$ and $A=2.5cm$ and we know that
$V_{max}=A\omega$
$\Rightarrow 10=2.5\omega$
$\Rightarrow \omega =4s^{-1}$
$T=\frac{2\pi }{\omega }=\frac{2\pi }{4}=\frac{3.14}{2}=1.57s$ $\Rightarrow$A correct
$a_{max}=A{\omega }^2=2.5\times 4^2=40cm{s}^{-2}$ $\Rightarrow$B correct
$v(x=1cm)=w\sqrt{A^2-x^2}=4\sqrt{{2.5}^2-1^2}=4\sqrt{5.25}=4\sqrt{\frac{525}{100}}=4\sqrt{\frac{21\times 25}{100}}=2\sqrt{21}cm/s$ $\Rightarrow$C correct & D wrong.