A 0.44 kg object connected to a light spring with a spring constant of 22.0 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest.
(a) Determine the maximum speed of the mass.
28.28 cm/s
(b) Determine the speed of the mass when the spring is compressed 1.5 cm.
26.22 cm/s
(c) Determine the speed of the mass when the spring is stretched 1.5 cm.
26.22 cm/s
(d) For what value of x does the speed equal one-half the maximum speed?
a,b,c are right. im just having trouble with (d)
When the speed is 1/2 of the maximum speed, the kinetic energy is 1/4 of the maximum value.
That means the potential energy is 3/4 of the total energy, which is also 3/4 of the maximum potential energy.
That in turn means that the deflection x is sqrt (3/4) or 0.866 of its maximum value, the 4 cm amplitude.
x = 3.464 cm
To determine the value of x when the speed equals one-half the maximum speed, you need to use the equation for the velocity of an object undergoing simple harmonic motion.
The equation for the velocity of an object connected to a spring undergoing simple harmonic motion is given by:
v = ω√(A^2 - x^2)
Where:
v = velocity of the object
ω = angular frequency of the motion
A = amplitude of the motion (maximum displacement from equilibrium position)
x = displacement from the equilibrium position
In this case, we are given the amplitude A as 4.0 cm (or 0.04 m). We can calculate the angular frequency ω using the formula:
ω = √(k/m)
Where:
k = spring constant = 22.0 N/m
m = mass of the object = 0.44 kg
Plugging in the values, we have:
ω = √(22.0 N/m / 0.44 kg) = √(50.0 rad/s^2) ≈ 7.071 rad/s
Now, let's substitute the values into the equation for velocity:
v = (7.071 rad/s)√(0.04^2 - x^2)
To find the value of x when the speed is half the maximum speed, we can set v equal to half of the maximum speed (v_max/2) and solve for x:
v_max/2 = (7.071 rad/s)√(0.04^2 - x^2)
We know from part (a) that the maximum speed (v_max) is 28.28 cm/s (0.2828 m/s), so:
0.2828 m/s / 2 = (7.071 rad/s)√(0.04^2 - x^2)
0.1414 m/s = (7.071 rad/s)√(0.0016 - x^2)
Now, we can square both sides of the equation to eliminate the square root:
(0.1414 m/s)^2 = [(7.071 rad/s)√(0.0016 - x^2)]^2
0.02 m^2/s^2 = (7.071 rad/s)^2 * (0.0016 - x^2)
0.02 = 50 * (0.0016 - x^2)
0.02 = 0.08 - 50x^2
Rearranging the equation gives:
50x^2 = 0.08 - 0.02
50x^2 = 0.06
x^2 = 0.06 / 50
x^2 = 0.0012
Taking the square root of both sides gives:
x = √(0.0012) ≈ 0.0346 m
Therefore, the value of x when the speed equals one-half the maximum speed is approximately 0.0346 m.