Question

A 280 g object is attached to a spring and executes simple harmonic motion with a period of 0.270 s. If the total energy of the system is 4.75 J. (a) Find the maximum speed of the object _______ m/s (b) Find the force constant of the spring_________ N/m (c) Find the amplitude of the motion_________m

Answer 1

Answer:

Explanation:

given,

mass of the object = 280 g = 0.28 kg

time period = 0.270 s

total energy of the system  = 4.75 J

                 $\dfrac{1}{2}\ m\ V^2 = 4.75 J$

maximum speed of the object V =     $\sqrt{ \dfrac{2 \times 4.75}{0.28} }$

                                              V= 5.82 m / s

(b)   force constant of the spring K = m ω²

where ω = angular frequency = 2π / T

          T= time period = 0.25 s

ω = 25.13 rad / s

K = 0.28 × 25.13²

K = 176.824 N / m

(c). Amplitude of motion A =    $\dfrac{V}{\omega}$

                                         =     $\dfrac{5.82}{25.13}$

A = 0.232 m