Question

1. a 1650 g mass is hung from a spring. if the spring vibrates 300 times in 8 minutes, what is the force constant of the spring? (25.4 n/m) 4 marks

Explanation:

Step1: Convert mass to kg and time to seconds

Mass \( m = 1650\ g = 1.65\ kg \)
Time for 300 vibrations is 8 minutes. Convert 8 minutes to seconds: \( 8\times60 = 480\ s \)
Period \( T \) (time for one vibration) is \( T=\frac{480}{300}=1.6\ s \)

Step2: Use the formula for the period of a mass-spring system \( T = 2\pi\sqrt{\frac{m}{k}} \)

Rearrange the formula to solve for \( k \):
First, square both sides: \( T^{2}=4\pi^{2}\frac{m}{k} \)
Then, solve for \( k \): \( k = \frac{4\pi^{2}m}{T^{2}} \)

Step3: Substitute the values of \( m \) and \( T \)

\( m = 1.65\ kg \), \( T = 1.6\ s \)
\( k=\frac{4\times\pi^{2}\times1.65}{(1.6)^{2}} \)
Calculate numerator: \( 4\times\pi^{2}\times1.65\approx4\times9.8696\times1.65\approx4\times16.28484\approx65.13936 \)
Denominator: \( (1.6)^{2}=2.56 \)
\( k=\frac{65.13936}{2.56}\approx25.4\ N/m \)

Answer:

The force constant \( k \) of the spring is approximately \( 25.4\ N/m \)