Answer to Question #184430 in Mechanics | Relativity for shy

Question

the elastic constant of a helical spring is 9 N/m. a 1 kg body is attached from the spring and is made to move with simple harmonic motion. determine the period of vibration of the body and the acceleration when the body is 40 cm from its equilibrium position if the amplitude is 90 cm

Answer 1

Explanations & Calculations

• When the spring-mass system comes to an equilibrium,

$\qquad\qquad \begin{aligned} \small T&=\small k\bar{x}=mg\\ \small \bar{x}&=\small \frac{mg}{k}=\frac{1kg\times9.8ms^{-2}}{9Nm^{-1}}=1.089 \,m \end{aligned}$

equilibrium position is located at a position where the spring stretches itself to 1.089m more.

• Applying Newton's second law downwards on the system for an excitement of $\small x$ downward,

$\qquad\qquad \begin{aligned} \small \downarrow F&=\small ma\\ \small mg-k(\bar{x}+x)&=\small m\ddot{x}\\ \small -kx&=\small m\ddot{x}\\ \small \ddot{x}&=\small -\frac{k}{m}\cdot x \cdots\cdots(1) \end{aligned}$

• Comparing this with the characteristic SHM equation: $\small \ddot{x}=-\omega^2 x$, period of vibration can be found.

$\qquad\qquad \begin{aligned} \small \omega&=\small\frac{2\pi}{T}= \sqrt{\frac{k}{m}}\\ \small T&=\small 2\pi\sqrt{\frac{m}{k}} \end{aligned}$

• Since the given distance of 40cm is measured with respect to the equilibrium position, simply using the equation (1), corresponding acceleration can be found.
• Since the equilibrium extension is greater than the given amplitude, this system can have a 40cm elongation in both the upper & bottom positions ($\small \pm 40cm$).
• Then,

$\qquad\qquad \begin{aligned} \small \ddot{x}&=\small -\frac{9Nm^{-1}}{1\,kg}\times (\pm 0.4\,m)\\ &=\small \mp\,3.6\,ms^{-2}\\ \small \ddot{x}&=\begin{cases} \small-3.6\,ms^{-2} \,@ \,x=+40\,cm(below\, the\, \bar{x}\\ \small +3.6\,ms^{-2}\,@\, x=-40\,cm(above\,the\, \bar{x} \end{cases} \end{aligned}$

$\qquad\qquad \begin{aligned} \small \end{aligned}$