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A 0.600-kg glider on an air track is attached to the end of an ideal spring with force constant 490 n/m ; it undergoes simple harmonic motion with an amplitude of 6.00��10-2m. What is the period of the motion?

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Final answer:

The period of the motion for a 0.600-kg glider attached to an ideal spring with a force constant of 490 N/m and undergoing simple harmonic motion is approximately 0.123 seconds.

Step-by-step explanation:

The period of the motion for a glider attached to an ideal spring undergoing simple harmonic motion (SHM) can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass of the glider, and k is the force constant of the spring.

Given a mass (m) of 0.600 kg and a force constant (k) of 490 N/m, we can substitute these values into the formula to find the period T:

T = 2π√(0.600 kg / 490 N/m)

After doing the calculation, we find that T ≈ 0.123 s.

Thus, the period of the motion for the 0.600-kg glider on the air track attached to the ideal spring is approximately 0.123 seconds.

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