Question

A particle of mass $0\cdot 50kg$ executes a simple harmonic motion under a force $F=-\left(50N/m\right)x$. If it crosses the centre of oscillation with a speed of $10m/s$, then the amplitude of the motion is

• A. $1m$
• B. $2m$
• C. $3m$
• D. $4m$

Answer 1

The correct option is A $1m$

A particle executing S.H.M acted upon by a force

F = -(50N/m)$x$

it mess m= 0.5kg

so, a = F/m = acceleration

$\Rightarrow a=-100x$

$\Rightarrow a+{\omega }^{2}x=0$

$\omega$ = angular frequency $=\sqrt{100}=10rad/s$

velocity at mean position

$\omega A=10/s$ A = amplitude of oscillation

$\Rightarrow A=\frac{10}{10}=1m$