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A mass 3 -kilogram attached to end of a spring that is stretched 20 cm by a force of 15 N. It is set in motion with initial position x(0)=0 and velocity v(0)=−10 m/s. Find the amplitude and the period of the resulting motion.

a. Amplitude =2m, freq. =5/2πHZ
b. Amplitude =2m, freq. =3/2πHZ
c. Amplitude =1 m, freq. =5/2πHZ
d. Amplitude =1m, freq. =3/2πHZ

User Kereen
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1 Answer

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Final answer:

The amplitude of the motion is 20 cm and the period is 2π/5√3 seconds.

Step-by-step explanation:

In this problem, we are given a mass of 3 kg attached to the end of a spring that is stretched 20 cm by a force of 15 N. The mass is set in motion with an initial position of x(0) = 0 and velocity of v(0) = -10 m/s. We need to find the amplitude and the period of the resulting motion.

The amplitude of the motion can be found using the formula A = x_max - x_eq, where x_max is the maximum displacement and x_eq is the equilibrium position. In this case, the equilibrium position is 20 cm, and the maximum displacement can be found using the initial position x(0) = 0. Therefore, the amplitude is 20 cm.

The period of the motion can be found using the formula T = 2π√(m/k), where m is the mass and k is the force constant of the spring. In this case, m = 3 kg and k = F/x_max = 15 N / 0.2 m = 75 N/m. Plugging in these values, we get T = 2π√(3/75) = 2π/5√3 s. Therefore, the period is 2π/5√3 seconds.