A 210 g block is attached to a horizontal spring andexecutes simple harmonic mot

Question

A 210 g block is attached to a horizontal spring andexecutes simple harmonic motion with a period of 0.350 s. If the total energy of the system is2.00 J, find the following (a) the force constant of the spring
1 N/m
(b) the amplitude of the motion.
2 m A 210 g block is attached to a horizontal spring andexecutes simple harmonic motion with a period of 0.350 s. If the total energy of the system is2.00 J, find the following (a) the force constant of the spring
1 N/m
(b) the amplitude of the motion.
2 m A 210 g block is attached to a horizontal spring andexecutes simple harmonic motion with a period of 0.350 s. If the total energy of the system is2.00 J, find the following (a) the force constant of the spring
1 N/m
(b) the amplitude of the motion.
2 m A 210 g block is attached to a horizontal spring andexecutes simple harmonic motion with a period of 0.350 s. If the total energy of the system is2.00 J, find the following A 210 g block is attached to a horizontal spring andexecutes simple harmonic motion with a period of 0.350 s. If the total energy of the system is2.00 J, find the following A 210 g block is attached to a horizontal spring andexecutes simple harmonic motion with a period of 0.350 s. If the total energy of the system is2.00 J, find the following A 210 g block is attached to a horizontal spring andexecutes simple harmonic motion with a period of 0.350 s. If the total energy of the system is2.00 J, find the following (a) the force constant of the spring
1 N/m
(b) the amplitude of the motion.
2 m (a) the force constant of the spring
1 N/m
(b) the amplitude of the motion.
2 m

Explanation / Answer

(a) Given T = 0.35 s
                 m = 210 g = 210 * 10-3 kg

w = (2/T) = (2 * 3.14/0.35) = 17.9 rad/s Since w = (k/m)1/2---> w2 = (k/m)---> k = w2 * m ---> k=(17.9)^2*210*10^(-3)=67.3 N/m


(b) Given E= 2J

Since A = (2E/k)1/2--->A = (2 * 2/67.3)1/2= 0.244 m = 244 * 10-3 m = 244mm

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