Question

A block of mass 0.404 0.404 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.666 0.666 m. Find the spring constant. N / m N/m The block is then pulled down an additional 0.359 0.359 m and released from rest. Assuming no damping, what is its period of oscillation? s s How high above the point of release does the block reach as it oscillates?

Answer 1

To solve this problem it is necessary to apply the concepts related to the Force from Hook's law as well as the definition of the period provided by the same definition.

We know that the Force can be defined as

$F = xk \rightarrow mg = kx \Rightarrow k = \frac{mg}{x}$

Where

k = Spring constant

x = Displacement

g = Gravity

m = mass

At the same time the period of a spring mass system is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Where

m = Mass

k = Spring constant

Our values are given as,

m = 0.404kg

x = 0.666m

Replacing to find the value of the Spring constant we have that

$k = \frac{mg}{x}$

$k = \frac{(0.404)(9.8)}{0.666}$

$k = 5.944N/m$

Now using the formula of the period we know that

$T = 2\pi \sqrt{\frac{m}{k}}$

$T = 2\pi \sqrt{\frac{0.404}{5.944}}$

$T = 1.638s$

Finally, if the oscillation was 0.359m

The maximum height will be determined by the total length of that oscillation being equivalent to

$h=2a$

$h = 2*0.359$

$h = 0.718m$