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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. 
6 months ago

Answers : (1)

Arun
20876 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)
6 months ago
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