Ncert Exercises & Solutions

14.24 A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is
(a) 5 cm (b) 3 cm (c) 0 cm..

It is given in the questions that:

$Amplitude, A = 5 cm = 0.05 m$
$Time period, T = 0.2 s$
$For displacement, x = 5 cm = 0.05 m$
$Acceleration :$
$a = - ω^{2} x$
$= - {(\frac{2 π}{T})}^{2} x$
$= - {(\frac{2 π}{0.2})}^{2} \times 0.05$
$= - 5 π^{2} m / s^{2}$
$Velocity :$
$v = ω \sqrt{A^{2} - x^{2}}$
$= \frac{2 π}{T} \sqrt{{(0.05)}^{2} - {(0.05)}^{2}}$
$= 0$
$For displacement, x = 3 cm = 0.03 m$
$Acceleration :$
$a = - ω^{2} x$
$= - {(\frac{2 π}{T})}^{2} x$
$= - {(\frac{2 π}{0.2})}^{2} = 0.03$
$= - 3 π^{2} m / s^{2}$
$Velocity :$
$\begin{array}{l} v = ω \sqrt{A^{2} - x^{2}} \\ = \frac{2 π}{T} \sqrt{A^{2} - x^{2}} \\ = \frac{2 π}{T} \sqrt{(0 .05)^{2} - (0 .03)^{2}} \\ = \frac{2 π}{0 .2} \times 0 .04 \\ = 0 .4 π m / s \end{array}$
$For displacement, x = 0$
$Acceleration :$
$a = - ω^{2} x = 0$
$Velocity :$
$\begin{aligned} v & = ω \sqrt{A^{2} - x^{2}} \\ = \frac{2 π}{T} \sqrt{A^{2} - x^{2}} \\ = & \frac{2 π}{0 .2} \sqrt{(0 .05)^{2} - 0} \\ = 0 .5 π m / s \end{aligned}$

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