Respuesta :
Answer:
Force constant of the spring (k) = 24.07 N/m
Concept/Theory:
The period [tex] \sf (T_s) [/tex] 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:
[tex] \boxed{ \bf{T_s = 2 \pi \sqrt{\dfrac{m}{k}}}}[/tex]
Explanation:
Mass = 0.250 kg
Period = 0.640 s
By substituting values in the equation, we get:
[tex] \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[/tex]
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 ([tex]k[/tex]), in newtons per meter, from the equation of the Period ([tex]T[/tex]), in seconds, which is described below:
[tex]T = 2\pi\cdot \sqrt{\frac{m}{k} }[/tex] (1)
Where [tex]m[/tex] is the mass of the moving element, in kilograms.
If we know that [tex]T = 0.640\,s[/tex] and [tex]m = 0.250\,kg[/tex], then the spring constant of the spring is:
[tex]0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }[/tex]
[tex]\sqrt{\frac{0.250}{k} } \approx 0.102[/tex]
[tex]\frac{0.250}{k} \approx 0.0104[/tex]
[tex]k \approx 24.038\,\frac{N}{m}[/tex]
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: https://brainly.com/question/17315536
