Question

A spring has an unstretched length of 14 cm . When an 81 g ball is hung from it, the length increases by 6.0 cm . Then the ball is pulled down another 6.0 cm and released.a. What is the spring constant of the spring?
b. What is the period of the oscillation?

Answer 1

To solve this problem we will apply the two concepts mentioned. To find the constant we will apply Hooke's law, and to find the period we will apply the relationship between the mass and the spring constant. Let us begin,

PART A) For this section we will use Hooke's law. In turn, since the force applied is equivalent to weight, we will use Newton's law for which weight is defined as the product between mass and gravity. This weight is equal to the Spring Force.

$F = k\Delta L$

Here,

k = Spring constant

$\Delta L$ = Displacement

F = Force, the same as the Weight (mg)

Then we have that

$mg = k\Delta L$

$k = \frac{mg}{\Delta L}$

$k = \frac{9.8*0.081}{0.06}$

$k = 13.23N/m$

Therefore the spring constant is 13.23N/m

PART B)  To find the period of oscillation, the relationship that allows us to find is given by the following mathematical function,

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

Here

m = mass

k = Spring constant

Replacing,

$T = 2\pi \sqrt{\frac{0.081}{13.23}}$

$T = 0.491s$

Therefore the period of the oscillation is 0.491s

Answer 2

(a) The spring constant of spring is 13.23 N/m

(b) The time period of oscillation is 0.49s

Spring constant and time period:

(a) The unstretched length of the spring is L = 14 cm = 0.14m

When the ball of mass m = 81 g = 0.081kg is hung from it, the stretch in the spring is x = 6 cm = 0.06m

The restoring force F of the spring balances the weight W of the ball:

F = W

kx = mg

where k is the spring constant

k = mg/x

k = 0.081×9.8/0.06

k = 13.23 N/m

(b) The time period of oscillation is defined as:

$T=2\pi\sqrt{\frac{m}{k} }\\\\T=2\pi\sqrt{\frac{0.081}{13.23} }$

T = 0.49s

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