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Grade: 12th pass
        
A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion. 
10 months ago

Answers : (1)

Arun
22557 Points
							
a) T = 2pi / w so w = 2 pi / T

w = sqrt ( k / m )

so k = w²* m 

b) (1/2)* k * xmax ² = 2,00 J

so, xmax = sqrt( 2,00 * 2 / k)
 
hopebit helps
 
Regards
Arun (askIITians forum expert)
10 months ago
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