Question

A toy is undergoing SHM on the end of a horizontal spring with force constant 308 N/m . When the toy is 0.130 m from its equilibrium position, it is observed to have a speed of 3 m/s and a total energy of 5.2 J .

Part A Find the mass of the toy.

Part B Find the amplitude of the motion.

Part C Find the maximum speed attained by the object during its motion.

Answer 1

part A :

apply the balancing condtion at equillibrium as

Energy U = KE = 0.5mv^2 + 0.5 kx^2 = 5.2 J

0.5 mv^2 = 5.2 -(0.5 * 308 * 0.13* 0.13)

0.5 mv^2 = 2.6

mass m = 2.6*2/(3*3)

mass m = 0.578kgs

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Part B:

0.5 kA^2 = 5.2

A^2 = 5.2 *2/308

A = 18.4cm

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Part C:

The maximum speed of the object is

Vmax=Aw

=A sqaure root (k/m) ;w=sqaure root (k/m)

=(0.184m) sqaure root (308/0.578)

=4.25 m/s