Question

A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant of the spring?

Answer 1

The force constant of the spring is approximately 24.038 newtons per meter.

As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant ($k$), in newtons per meter, from the equation of the Period ($T$), in seconds, which is described below:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$ (1)

Where $m$ is the mass of the moving element, in kilograms.

If we know that $T = 0.640\,s$ and $m = 0.250\,kg$, then the spring constant of the spring is:

$0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }$

$\sqrt{\frac{0.250}{k} } \approx 0.102$

$\frac{0.250}{k} \approx 0.0104$

$k \approx 24.038\,\frac{N}{m}$

The force constant of the spring is approximately 24.038 newtons per meter.

Please see this question related to Simple Harmonic Motion for further details: brainly.com/question/17315536

Answer 2

Answer:

Force constant of the spring (k) = 24.07 N/m

Concept/Theory:

The period $\sf (T_s)$ of a spring-mass system is proportional to the square root of the mass (m) and inversely proportional to the square root of the force constant of the spring (k).

Equation of period:

$\boxed{ \bf{T_s = 2 \pi \sqrt{\dfrac{m}{k}}}}$

Explanation:

Mass = 0.250 kg

Period = 0.640 s

By substituting values in the equation, we get:

$\rm \longrightarrow 0.640 = 2 \pi \sqrt{\dfrac{0.250}{k}} \\ \\ \rm \longrightarrow 2 \times 3.14 \sqrt{ \dfrac{0.25 0}{k} } = 0.640 \\ \\ \rm \longrightarrow 6.28 \sqrt{ \dfrac{0.250}{k} } = 0.640 \\ \\ \rm \longrightarrow \sqrt{ \dfrac{0.250}{k} } = \frac{0.640}{6.28} \\ \\ \rm \longrightarrow \frac{0.250}{k} = { \bigg(\frac{0.640}{6.28} \bigg) }^{2} \\ \\ \rm \longrightarrow \frac{k}{0.250} = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \\ \\ \rm \longrightarrow k = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \times 0.250 \\ \\ \rm \longrightarrow k = 24.07 \: N/m$