What is the maximum velocity of a spring oscillator with a mass of 0.7 kg and a spring constant of 74 N/m that is released from rest at a position of x = 0.1 m?
max KE=maxspringPE
1/2 m v^2=1/2 k x^2
solve for v.
I dont know please tell me?
To determine the maximum velocity of the spring oscillator, we need to use the equations of motion for a simple harmonic oscillator. The equation that relates the maximum velocity (Vmax) with the mass (m) and the spring constant (k) is given by:
Vmax = Aω
Where:
A is the amplitude of the oscillation, which is the maximum displacement from the equilibrium position, and
ω is the angular frequency of the oscillation, which is given by ω = √(k / m)
To find the amplitude (A) of the oscillation, we need to know the initial position of the oscillator (x0). In this case, we are given that the oscillator is released from rest at a position of x0 = 0.1 m.
So, the amplitude (A) is equal to the absolute value of the initial position (|x0|). Therefore, A = |0.1 m| = 0.1 m.
Now, we can substitute the values of the mass (m) and the spring constant (k) into the angular frequency equation to calculate ω:
ω = √(k / m)
= √(74 N/m / 0.7 kg)
= √(106.43 N/kg)
≈ 10.317 rad/s (rounded to three decimal places)
Finally, we can substitute the amplitude (A) and the angular frequency (ω) into the equation for the maximum velocity (Vmax) to calculate the answer:
Vmax = Aω
= 0.1 m × 10.317 rad/s
≈ 1.032 m/s (rounded to three decimal places)
Therefore, the maximum velocity of the spring oscillator is approximately 1.032 m/s.