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Grade: 11
        
A 5.13-kg object moves on a horizontal frictionless surface under the influence of a spring with force constant 9.88 N/cm. The object is displaced 53.5 cm and given an initial velocity of 11.2 m/s back toward the equilibrium position. Find (a) the frequency of the motion, (b) he initial potential energy of the system, (c) the initial kinetic energy, and (d) the amplitude of the motion.
2 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
							235-483_12.PNG
235-1251_13.PNG
= √(2×463 J)/(9.88 N/cm)
= √(2×463 J)/(9.88 N/cm×1 cm/10-2 m)
= 0.968 √J / N/m
= 0.968 √(J.m / N) (1 N.m/ 1J)
=0.968 m
From the above observation we conclude that, the amplitude xm of the motion would be 0.968 m.
2 years ago
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