Question

question 3 (1 point)
a 0.250-kg stone is attached to an ideal spring and undergoes simple harmonic oscillations with a period of 0.640 s. what is the force constant (spring constant) of the spring?

a) 2.45 n/m

b) 0.102 n/m

c) 12.1 n/m

d) 0.610 n/m

e) 24.1 n/m

Explanation:

Step1: Recall the formula for the period of a mass-spring system

The formula for the period \( T \) of a mass \( m \) attached to a spring with spring constant \( k \) is \( T = 2\pi\sqrt{\frac{m}{k}} \). We need to solve this formula for \( k \).

First, square both sides of the equation: \( T^{2}=4\pi^{2}\frac{m}{k} \).

Then, rearrange the formula to solve for \( k \): \( k=\frac{4\pi^{2}m}{T^{2}} \).

Step2: Substitute the given values into the formula

We are given that \( m = 0.250\space kg \) and \( T = 0.640\space s \).

Substitute these values into the formula for \( k \):

\( k=\frac{4\pi^{2}\times0.250}{(0.640)^{2}} \)

First, calculate the numerator: \( 4\pi^{2}\times0.250 = \pi^{2}\times1\approx9.8696 \)

Then, calculate the denominator: \( (0.640)^{2}=0.4096 \)

Now, divide the numerator by the denominator: \( k=\frac{9.8696}{0.4096}\approx24.1\space N/m \)

Answer:

e) 24.1 N/m