a mass of 0.7kg is attached to a horizontal spring (k=5.2) the spring is compressed to 20cm and released . assume that the spring is ideal and frictionless and the mass oscillates horizontally, 1.what is the amplitude of oscillation
2.what is the maximum speed
3.what are the speed and acceleration of the mass when its displacement is at 12cm
1.
Max. compression is 20cm - so the amplitude A is 0.2m.
2.(1/2)mV^2 = (1/2)kA^2
V^2 = kA^2/m = 5.2*0.04/0.7
V=0.54m/s
3.Speed at X = sqrt(k/m)sqrt(A^2-X^2)
=2.72sqrt(.04-0.0144)
= 2.72*0.16= 0.43m/s
Acc. at X= (k/m)x= (5.2/0.7)*0.12
=0.89 m/s^2
To answer these questions, we need to use the equations of motion for a spring-mass system. Let's go through each question step by step:
1. Amplitude of Oscillation:
The amplitude of oscillation can be determined by the maximum displacement from the equilibrium position. In this case, the spring is initially compressed to 20 cm, so when released, the mass will move to an equal distance in the opposite direction. Therefore, the amplitude of oscillation is given by the initial displacement, which is 20 cm.
2. Maximum Speed:
The maximum speed occurs at the equilibrium position (x = 0), where the spring's potential energy is fully converted to kinetic energy, and the mass has maximum velocity. We can calculate the maximum speed using the equation:
v_max = ω * A
where ω is the angular frequency and A is the amplitude of oscillation. The angular frequency (ω) can be calculated using the following formula:
ω = sqrt(k/m)
where k is the spring constant and m is the mass. Plugging in the given values:
ω = sqrt(5.2 / 0.7) = 2.63 rad/s
Now we can substitute the values back into the initial equation to find the maximum speed:
v_max = 2.63 * 0.2 = 0.526 m/s
So, the maximum speed is 0.526 m/s.
3. Speed and Acceleration at 12 cm displacement:
To find the speed and acceleration at a specific displacement, we need to use the equation of motion for a spring-mass system:
x(t) = A * cos(ωt + φ)
where x(t) is the displacement at time t, A is the amplitude, ω is the angular frequency, and φ is the phase constant. At t = 0, the mass is at maximum displacement, so the equation becomes:
x(t) = A * cos(φ)
Given that x(t) = 12 cm, we can solve for the phase constant φ:
12 cm = 0.2 m * cos(φ)
cos(φ) = 0.6
φ = arccos(0.6) ≈ 0.927 rad
Now, we differentiate the equation of motion with respect to time to find the velocity:
v(t) = -A * ω * sin(ωt + φ)
At the displacement of 12 cm, the equation becomes:
v(t) = -0.2 m * 2.63 rad/s * sin(2.63t + 0.927)
At t = 0, the velocity can be found:
v(t=0) = -0.2 m * 2.63 rad/s * sin(0.927) ≈ -0.181 m/s
Therefore, at a displacement of 12 cm, the speed is approximately 0.181 m/s.
To find the acceleration, we differentiate the velocity equation with respect to time:
a(t) = -A * ω^2 * cos(ωt + φ)
At the displacement of 12 cm, the equation becomes:
a(t) = -0.2 m * (2.63 rad/s)^2 * cos(2.63t + 0.927)
At t = 0, the acceleration can be found:
a(t=0) = -0.2 m * (2.63 rad/s)^2 * cos(0.927) ≈ -3.52 m/s^2
Therefore, at a displacement of 12 cm, the acceleration is approximately -3.52 m/s^2.
In summary:
1. The amplitude of oscillation is 20 cm.
2. The maximum speed is 0.526 m/s.
3. At a displacement of 12 cm, the speed is approximately 0.181 m/s and the acceleration is approximately -3.52 m/s^2.