Answer:
the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s
Explanation:
Given that;
mass of the object m = 1.20 kg
period of oscillation = 0.750 s
Amplitude ( A/x) = 15.0 cm = 0.15 m
now;
a) Determine the oscillation frequency;
oscillation frequency f = 1/T
we substitute
f = 1 / 0.750 s
f = 1.33 Hz
Therefore, the oscillation frequency is 1.33 Hz
b) Determine the spring constant;
we solve for spring constant from the following expression;
T = 2π√(m/k)
k = 4π²m / T²
so we substitute
k = (4π² × 1.20) / (0.750)²
k = 47.3741 / 0.5625
k = 84.22 N/m
Therefore, the spring constant is 84.22 N/m
c) determine the speed of the mass when it is halfway to the equilibrium position
So, at equilibrium, the energy is equal to K.E
such that;
1/2mv² = 1/2kx²
mv² = kx²
v² = kx² / m
v = √( kx²/m)
we substitute
v = √( 84.22×(0.15 m)²/ 1.2 )
v = √( 1.89495 / 1.2 )
v = √ 1.579125
v = 1.26 m/s
Therefore, the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s
