Find Max Speed of 3.4g Mass in Oscillatory Motion: Energy Conservation

[map7s]

There is a displacement versus time graph of a 3.4 g mass on a spring that is in oscillatory motion. A=0.5 and the wavelength for one period is 3 s while the total time shown on the graph is 7 s. I need to find out the maximum speed of this mass.

At first I thought that it would easily be the amplitude divided by the time at that highest amplitude, but obviously that was wrong. I then decided to try energy conservation, which I think is right, but I think that I am not using the correct equation. I've been trying to do some form of mgh=1/2 mv^2 and solving for v. I changed h to x and I knew that x=Acos((2pi/T)t) so I put that in for h and I tried to solve for v. I plugged in my answer and it was wrong, so I'm pretty sure that I just didn't use a correct form of this equation. How can I get a more appropriate form of this equation to use (if it is indeed the correct method) ?

 

[OlderDan]

If this is a calculus based course, you know the velocity is the derivative of the displacement. If you can write x(t) you can find v(t). If it is not calculus based, almost surey your text gives you the equations for displacement, velocity, and acceleration of a harmonic oscillator as a function of time. From the graph you can find the things you need to use those equations.

 

[kyle8921]

It is correct to use energy conservation to find the maximum speed of the 3.4 g mass in oscillatory motion. However, the equation you have used, mgh=1/2 mv^2, is for objects in a gravitational field and does not apply to objects in oscillatory motion. The correct equation to use is the conservation of mechanical energy, which states that the total mechanical energy (potential energy + kinetic energy) remains constant throughout the motion.

In this case, the potential energy is given by the equation U = 1/2 kx^2, where k is the spring constant and x is the displacement from equilibrium. The kinetic energy is given by the equation K = 1/2 mv^2, where m is the mass and v is the velocity.

Since the mass is in simple harmonic motion, the total mechanical energy will be the sum of the potential and kinetic energies at any given point in time. Therefore, at the maximum displacement (A), the total mechanical energy will be equal to the potential energy, and at the equilibrium point (x=0), the total mechanical energy will be equal to the kinetic energy.

Setting these two equations equal to each other, we can solve for the maximum speed (vmax) at the amplitude:

1/2 kA^2 = 1/2 mvmax^2

Solving for vmax, we get:

vmax = A√(k/m)

Plugging in the given values, we get:

vmax = (0.5)(0.034m)(√(k/0.0034kg))

To find the value of k, we can use the equation k = 4π^2m/T^2, where T is the period of oscillation. Plugging in the values, we get:

k = 4π^2(0.0034kg)/(3s)^2 = 0.474 N/m

Substituting this value for k in the previous equation, we get:

vmax = (0.5)(0.034m)(√(0.474 N/m /0.0034kg)) = 0.53 m/s

Therefore, the maximum speed of the 3.4 g mass in oscillatory motion is 0.53 m/s.

It is important to note that this calculation assumes that there is no energy loss due to friction or other factors. In real-world situations, there may be