Question

A 0.158 kg toy is undergoing SHM on the end of a horizontal spring with force constant 310 N/m . When the object is 1.50×10⁻² m from its equilibrium position, it is observed to have a speed of 0.320 m/s .

A: Find the total energy of the object at any point in its motion.

B: Find the amplitude of the motion.

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

Answer 1

Mass m = 0.158 kg

Force constant of the spring k = 310 N/m

Angular frequency = [k/m]

                               = 44.29 rad/s

Position x = 1.5 x10 ^-2 m = 0.015 m

Speed v = 0.32 m/s

We know v = [A ² - x ² ]

        0.32 = 44.29 [A ² - (0.015) ² ]

[A ² - (0.015) ² ]= 7.22 x10 ^-3

[A ² - (0.015) ² ] = 5.219 x10 ^-5

                A ² = 2.77 x10 ^-4

                   A = 0.01664

                      = 1.664 x10 ^-2 m

(A).the total energy of the object at any point in its motion = (1/2) kA 2

                                                                                     = 0.5 x 1.664 x10 ^-2 x 310 ²

                                                                                     = 800 J

B: the amplitude of the motion.A = 1.664 x10 ^-2 m

C: the maximum speed attained by the object during its motion = A

                                                                                           = 0.736 m/s