Question

A block of mass 0.464 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.646 m. Find the spring constant. N/m The block is then pulled down an additional 0.338 m and released from rest. Assuming no damping, what is its period of oscillation? s How high above the point of release does the block reach as it oscillates?

Answer 1

Explanation:

Mass, m = 0.464 kg

Compression in the spring, x = 0.646 m

(a) The net force acting on the spring is given by :

$kx=mg$

k is spring constant

$k=\dfrac{mg}{x}\\\\k=\dfrac{0.464\times 9.8}{0.646 }\\\\k=7.03\ N/m$

(b) The angular frequency of the spring mass system is given by :

$\omega=\sqrt{\dfrac{k}{m}} \\\\\omega=\sqrt{\dfrac{7.03}{0.464 }} \\\\\omega=3.89\ rad/s$

The period of oscillation is :

$T=\dfrac{2\pi}{\omega}\\\\T=\dfrac{2\pi}{3.89}\\\\T=1.61\ m/s$

(c) As the spring oscillates, its will reach to a height of 2(0.338) = 0.676 m

Answer 2

(a) The spring constant of the spring is 7.04 N/m.

(b) The angular frequency of the spring is 3.89 rad/s.

(c) The period of the oscillation is 1.62 s.

The given parameters;

• mass of the block, m = 0.464 kg
• extension of the spring, x = 0.646 m
• additional extension, = 0.338 m

The spring constant of the spring is calculated as follows;

$F = mg\\\\kx = mg\\\\k = \frac{mg}{x} \\\\k = \frac{0.464 \times 9.8}{0.646} \\\\k = 7.04 \ N/m$

The angular frequency of the spring is calculated as follows;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{7.04}{0.464} } \\\\\omega = 3.89 \ rad/s$

The period of the oscillation is calculated as follows;

$T = \frac{2\pi}{\omega} \\\\T = \frac{2 \pi }{3.89} \\\\T = 1.62 \ s$

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