Question

A block of mass 0.427 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result the spring is stretched by 0.612 m. Find the spring constant Number N/m The block is then pulled down an additional 0.317 m and released from rest. Assuming no damping, what is its period of oscillation? Number How high above the point of release does the block reach as it oscillates? Number In

Answer 1

The spring constant is

The time period is

$T = 2\pi\sqrt{\frac{m}{k}} = 6.28\times \sqrt{\frac{0.427}{6.84}} = 1.57s$

Answer 2

The spring constant is defined as

$$k=\frac{F}{\mathrm{\Delta }y}$$

where ?$F$ denotes the stretching (or compressing) force and Δ?$\mathrm{\Delta }y$ is the resulting displacement of the end of the spring. In the present case, the stretching force is the weight of the block,

$$F=mg$$

where ?$m$ is the block's mass and ?$g$ denotes the acceleration due to gravity. Substitute this in the formula for the spring constant to obtain the result in terms of the given quantities.

$$k=\frac{mg}{\mathrm{\Delta }y}$$

Now substitute the numerical data to obtain the numerical value of the spring constant.

$$k=\frac{\left(0.466\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)}{0.668\text{ m}}=6.84\text{ N/m}$$

The formula for the period of oscillation is

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

Substitute the formula for the spring constant in terms of given quantities, which you obtained previously, and cancel the mass to obtain the period in terms of given quantities.

$$T=2\pi \sqrt{\frac{\mathrm{\Delta }y}{g}}$$

Substitute the given values.

$$T=2\pi \sqrt{\frac{0.668\text{ m}}{9.81{\text{ m/s}}^2}}=1.64\text{ s}$$

After its release from rest, the block oscillates about its equilibrium position in simple harmonic motion with an amplitude ?$A$, which equals the distance from which it was released below the equilibrium position. Therefore, it reaches the same distance, ?$A$, above the equilibrium position as it was released below the equilibrium position. In other words, it reaches a height of 2?$2A$ above its release point.

2?=2×(0.382 m)=0.764 m