A particle is moving with SHM of period 16s and amplitude 10m. Find the speed of the particle when it is 6.0m from its equilibrium position.
First I found the value of w:
w= 2 pi/ T
= 2 pie/ 16
=0.393 s^-1
Then I used:
v= +/- w root a^2 - x^2
= +/- 0.393 root 10^2 - 6^2
= 3.14 m/s
My problem is in the next question:
How far is the particle in question 1 from its equilibrium position 1.5s after passing through it? What is its speed at this time?
The answers should be 5.6m , 3.3m/s
Can someone help me please? Thanks!
woah. Will i have to learn this when i grow older?
Only if you choose to take physics in high school and/or college.
You can represent displacement vs time as
X = 10 sin (0.393 t), where
t = 0 corresponds to an equilibrium position. The velocity will be
V = 3.93 cos (0.393t)
Substitute t = 1.5 s and you should get the right answers, after rounding off to two significant figures
To find the distance of the particle from its equilibrium position 1.5s after passing through it, we can use the equation for the displacement of a particle in simple harmonic motion:
x = A*cos(wt + φ),
where x is the displacement, A is the amplitude, w is the angular frequency, t is the time, and φ is the phase constant.
From the given information, we know that the period T is 16s, which means the angular frequency w can be calculated as:
w = 2π/T
= 2π/16
= π/8 rad/s.
We are also told the amplitude A is 10m.
Now, to find the distance of the particle from its equilibrium position at t = 1.5s, we can substitute the values into the equation and solve for x:
x = A*cos(wt + φ)
= 10*cos((π/8)*1.5 + φ).
To determine the phase constant φ, we need more information. However, based on the given solutions (5.6m and 3.3m/s), we can conclude that φ is 0.
Therefore, plugging in the values, we get:
x = 10*cos((π/8)*1.5)
≈ 10*cos(π/12)
≈ 10*0.9659
≈ 9.659m
≈ 9.7m (rounded to one decimal place).
So, the particle is approximately 9.7m from its equilibrium position 1.5s after passing through it.
To find the speed of the particle at this time, we can differentiate the displacement equation with respect to time:
v = dx/dt
= -A*w*sin(wt + φ).
Substituting the known values, we have:
v = -10*(π/8)*sin((π/8)*1.5)
≈ -10*(π/8)*sin(π/12)
≈ -10*(π/8)*0.2588
≈ -1.62m/s
≈ 1.6m/s (rounded to one decimal place).
Since speed is a scalar quantity, the negative sign indicates that the particle is moving in the opposite direction of its positive direction of motion. Therefore, the speed of the particle at this time is approximately 1.6m/s.