Question

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again?

Answer 1

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$, $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$, then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$, $k = 62.237\,\frac{N}{m}$, $m=1.65\,kg$, $x(t) = 0\,m$ and $\phi = 0\,rad$, then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$. Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.