A 200 g air-track glider is attached to a spring. The glider is pushed in 9.8 cm against the spring, then released. A student with a stopwatch finds that 12 oscillations take 15.0 s. Part A What is the spring constant

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Answer:

k =  5.05 N/m

Explanation:

In order to calculate the spring mass of the system, you use the following formula:

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]     (1)

T: period of oscillation of the system

m: mass of the air-track glider = 200g = 0.200 kg

k: spring constant = ?

You first calculate the period of oscillation:

[tex]T=\frac{1}{f}=\frac{1}{12/15.0s}=1.25s[/tex]

Next, you solve the equation (1) for k, and then you replace the values of the other parmateres:

[tex]k=4\pi^2 \frac{m}{T^2}\\\\k=4\pi^2 \frac{0.200kg}{(1.25s)^2}=5.05\frac{N}{m}[/tex]

The spring constant of the spring is 5.05 N/m

The spring constant of the spring is 5.1N/m

The formula for calculating the period of oscillation is expressed as:

[tex]T=2 \pi\sqrt{\frac{m}{k} }[/tex]

m is the mass of the spring

k is the spring constant

Making the spring constant "k" the subject of the formula will give;

[tex]k=4\pi^2\frac{m}{T^2}[/tex]

Get the period of oscillation "T"

[tex]T = \frac{1}{f}[/tex]

frequency "f" is the number of oscillations completed in one second.

If a student with a stopwatch finds that 12 oscillations take 15.0 s, the number of oscillations in one sec will be 15/12 = 1.25 osc.

Period T = 1/1.25 = 0.8sec

Get the required spring constant

[tex]k=4\pi^2\frac{m}{T^2}\\k=4(3.14)^2\frac{0.2}{0.8^2}\\k=5.1N/m[/tex]

Hence the spring constant of the spring is 5.1N/m

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