Spring/block on horizontal plane SHM

[physninj]

Homework Statement

1. Find the speed and acceleration when the kinetic energy is equal to half the potential energy
2. Find the minimum displacement between the points where the kinetic energy and acceleration are at half their maximum values

Homework Equations

mv^2=.5kx^2

The Attempt at a Solution

My correct harmonic equations

x(t)=(√2/10)cos(10t+5∏/4)
v(t)=-(√2)sin(10t+5∏/4)
a(t)=-(10*√2)cos(10t+5∏/4)

Got down to for part 1:
arctan(√(1/2))=10t+5∏/4

All I want to know is if its acceptable to use the negative time value that comes from not shifting the output of the inverse tangent, to plug in for velocity and acceleration. Personally I don't see why not.

And for part 2...what the bloody heck are they asking? I suppose I could find the displacements for each of those points and see when they get closest...I don't understand the goal of such a method though.

If you want to see the whole problem I have attached it. Thank you for any help.

 

[haruspex]

for 1, seems fine to me.
For 2, yes, there will be an infinite set of points at which these conditions arise, but there'll be a minimum proximity. Probably something like pi*n*apha+beta and pi*n*alpha+gamma, and you just have to figure out how close such points can get.

 

[physninj]

So is there some way to set up an equation for those displacements, and use a derivative set to zero to find a minimum? Thats the idea I am having right meow anyways

 

[haruspex]

So is there some way to set up an equation for those displacements, and use a derivative set to zero to find a minimum? Thats the idea I am having right meow anyways

No, you won't use differentiation. It's not minimum in a continuous function. It'll be like sin x = .5; x = π/6, 5π/6, 13π/6, ... Smallest difference = 4π/6.

 

[Nickie]

I would like to provide a response to the given content.

1. To find the speed and acceleration when the kinetic energy is equal to half the potential energy, we can use the equation mv^2 = 0.5kx^2, where m is the mass, v is the velocity, k is the spring constant, and x is the displacement. Rearranging this equation, we get v = ±√(kx^2/2m). Since the kinetic energy is equal to half the potential energy, we can substitute 0.5kx^2 for mv^2, which gives us v = ±√(kx^2/4m). Similarly, we can obtain the acceleration by differentiating the equation twice, which gives us a = ±(kx/m). Therefore, the speed and acceleration when the kinetic energy is equal to half the potential energy is given by v = ±√(kx^2/4m) and a = ±(kx/m), respectively.

2. To find the minimum displacement between the points where the kinetic energy and acceleration are at half their maximum values, we need to find the maximum values of kinetic energy and acceleration. The maximum kinetic energy will occur at the equilibrium position (x = 0), where the displacement is zero. Therefore, the maximum kinetic energy is given by 0.5k(0)^2 = 0. Similarly, the maximum acceleration will occur at the maximum displacement, which is given by the amplitude of oscillation. Therefore, the maximum acceleration is given by kA/m, where A is the amplitude. Now, we need to find the minimum displacement between the points where the kinetic energy and acceleration are at half their maximum values. This can be done by setting the kinetic energy and acceleration equations equal to half their maximum values and solving for x. So, we have 0.5kx^2 = 0.5kA^2/4 and kx/m = kA/2m. Solving these equations, we get x = ±A/√2. Therefore, the minimum displacement between the points where the kinetic energy and acceleration are at half their maximum values is given by ±A/√2.

In conclusion, it is acceptable to use the negative time value obtained from the inverse tangent to plug in for velocity and acceleration. And for part 2, the goal is to find