A 300g particle oscillating in SHM travels 35cm between the two extreme points in its motion with an average speed of 30cm/s. Find:
a) The angular frequency.
b) The maximum force on the particle.
b) The maximum speed.
.35 meters in half a period, T/2
2 A = .35
x = A sin 2 pi f t
x = A sin (2 pi t/T)
max at T/4
min at 3T/4
3T/4 - T/4 = time between max and min = T/2
so
0.30 m/s = 0.35 m /(T/2)s
T/2 = 0.35/.3
T = 7/3 seconds
w = ang freq = 2pi/T = 6 pi/7
max a = w^2 A = (6pi/7)^2(.35/2)
max v = w A you do it
To find the answers to these questions, we need to use the formulas for Simple Harmonic Motion (SHM). Here's how we can approach each question:
a) The angular frequency (ω) is related to the period (T) of the motion through the formula ω = 2π/T. Since we are given the average speed (v_avg), we can find the period using the formula T = (2d)/v_avg, where d is the distance between the two extreme points. Let's calculate it:
T = (2 * 0.35m) / 0.3m/s = 2.3333 s (approx)
Now, we can find the angular frequency:
ω = 2π / T = 2π / 2.3333s ≈ 2.69 rad/s
So, the angular frequency is approximately 2.69 rad/s.
b) The maximum force (F_max) on the particle in SHM occurs when it is at the extreme points. The formula for the maximum force is F_max = mω^2A, where m is the mass of the particle and A is the amplitude of the motion (half the distance traveled between extreme points).
Given:
m = 300g = 0.3kg
A = d/2 = 0.35m/2 = 0.175m
Substituting these values into the formula:
F_max = 0.3kg * (2.69 rad/s)^2 * 0.175m = 0.81 N (approx)
Therefore, the maximum force on the particle is approximately 0.81 N.
c) The maximum speed (v_max) of the particle occurs at the equilibrium position, where its displacement is zero. The formula for the maximum speed is v_max = ωA.
Given:
ω = 2.69 rad/s
A = d/2 = 0.35m/2 = 0.175m
Substituting these values into the formula:
v_max = 2.69 rad/s * 0.175m ≈ 0.47 m/s
So, the maximum speed of the particle is approximately 0.47 m/s.