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Chapter 1: Q8E (page 460)

In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring鈥檚 force constant.

Short Answer

Expert verified

The value spring鈥檚 force constant is 0.292 N/m.

Step by step solution

01

Calculate time period and angular frequency of oscillation

It is given that 0.2 kg air-track glider is attached to the end of an ideal spring of negligible mass and stars oscillation.

In a simple harmonic motion, the time taken to go from equilibrium to maximum amplitude and then again back to equilibrium is half time period of the oscillation,

Hence, the time period of oscillation will be,

T=2tT=2脳2.6T=5.2s

From T calculate angular frequency,

蝇=2鈭�T蝇=2鈭�5.2s蝇=1.21rad/s

02

Determine the value of K

We know that,

蝇=km

Where is equal to the angular frequency, m is the mass of the object and K is the spring鈥檚 force constant.

k=0.2kg脳1.21rad/s2k=0.292kg/s2k=0.292N/m

Hence, the value of spring鈥檚 force constant K=0.292 N/m.

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