a 2 kg frictionless block is attached to an ideal spring with force constant 300N/m. At t=0, the spring is neither stretched or compressed and the block is moving in the negative direction at 12 m/s. find (a) amplitude, (B)phase angle (c) write an equation for the position as a function of time.
I got a and b. For the amplitude, it is 0.98 m.
For the phase angle, I got 90 degrees or pi/2
C) I know the velocity equation is
v(t)=-wAsin(wt+phaseangle)
x(t)=Acos(wt +phaseangle)
My answer was x(t)=0.98cos(12.23t+pi/2)
The way I got the angular acceleration was using w=sqt(k/m)
HOWEVER in the book, the answer was
cos (ωt+(π/2))=−sin ωt, so x=(−0.98 m) sin((12.2 rad/s)t).
WHY IS IT SINE AND NOT COSINE?
you can write it as x=-.98cos(12.2t+PI/2)
That is the same as x=-.98sin(12.2t)
This is using the trig identity given
cos(wt+PI/2)=-sinwt
The equation for the position of an object attached to an ideal spring is given by:
x(t) = A * cos(ωt + φ)
Where:
- x(t) is the position of the object at time t,
- A is the amplitude of the motion,
- ω is the angular frequency, given by ω = √(k/m),
- t is the time, and
- φ is the phase angle.
In this case, you correctly found the amplitude (A) to be 0.98 m and the angular frequency (ω) to be 12.23 rad/s. However, the phase angle (φ) should be π/2 (or 90 degrees), as you mentioned.
Therefore, the correct equation for the position as a function of time is:
x(t) = 0.98 * cos(12.23t + π/2)
Both cosine and sine functions are periodic and have the same shape, but they differ by a phase shift of π/2. So, the solution given in the book using sine is mathematically equivalent to your solution using cosine:
cos(ωt + π/2) = -sin(ωt)
Thus, the answer x(t) = (-0.98 m) * sin((12.23 rad/s)t) is correct and equivalent to your answer x(t) = 0.98 * cos(12.23t + π/2). Both expressions represent the same motion.
The equation for the position of the block attached to the spring is given by:
x(t) = A cos(ωt + Φ)
where A is the amplitude, ω is the angular frequency, t is time, and Φ is the phase angle.
In this case, you correctly found the amplitude A to be 0.98 m and the phase angle Φ to be π/2 or 90 degrees. However, the book's answer is in terms of sine instead of cosine.
The reason for this difference is that it's just a matter of convention and how we choose to define the starting point for the motion. In this context, it is common to consider the equilibrium position of the block (neither compressed nor stretched) as the reference point for time t = 0.
When t = 0, the position x(t) should be equal to the amplitude A. By using cosine in the equation, we would have:
x(0) = A cos(ω * 0 + Φ)
= A cos(Φ)
= A
This means that when t = 0, the block is at the amplitude of the oscillation. However, the problem states that at t = 0, the block is moving in the negative direction at 12 m/s. This implies that the equilibrium position is shifted by a phase angle of π/2 or 90 degrees and the block is initially moving towards the negative direction.
To match this initial condition, we need to use a sine function instead of cosine. The sine function is shifted by π/2 or 90 degrees from the cosine function:
sin(ωt + Φ) = cos(ωt + Φ + π/2)
Therefore, the equation for the position of the block becomes:
x(t) = A sin(ωt + Φ)
= -0.98 m sin((12.23 rad/s)t)
So, both the book's answer and your answer are correct, just representing the equation in different forms based on the initial conditions given in the problem.