A 300g particle oscillating in SHM travels 11cm between the two extreme points in its motion with an average speed of 65 cm/s.
a) Find the angular frequency
b)The maximum force on the particle
c)The maximum speed
a) The formula for the angular frequency (ω) in SHM is given by ω = 2πf, where f is the frequency. We can find the frequency by dividing the average speed (v_avg) by the distance traveled (s), and then multiply it by 2π.
First, let's convert the distance traveled from centimeters to meters:
s = 11 cm = 0.11 m
Next, we can find the frequency (f) using the formula:
f = v_avg / s
f = 65 cm/s / 0.11 m
f ≈ 590.91 Hz
Finally, we can find the angular frequency (ω):
ω = 2πf
ω = 2π * 590.91 Hz ≈ 3708.29 rad/s
b) The maximum force (F_max) on the particle in SHM can be found using the formula F_max = mω²A, where m is the mass and A is the amplitude (half the distance traveled).
m = 300 g = 0.3 kg
A = s / 2 = 0.11 m / 2 = 0.055 m
F_max = (0.3 kg) * (3708.29 rad/s)² * (0.055 m)
F_max ≈ 3.6 N
c) The maximum speed (v_max) of the particle in SHM can be found by multiplying the amplitude (A) with the angular frequency (ω).
v_max = A * ω
v_max = 0.055 m * 3708.29 rad/s
v_max ≈ 203.46 m/s
Please note that these calculations assume ideal conditions and provide theoretical values.
a) To find the angular frequency, we can use the equation:
angular frequency = 2π / period
The period of the oscillation is the time it takes for the particle to complete one full cycle. Since the particle travels 11 cm between the two extreme points, it covers a total distance of 22 cm in one full cycle.
Therefore, the period can be calculated as:
period = distance / average speed = 22 cm / 65 cm/s
First, we need to convert the values to consistent units. Let's convert cm to meters:
period = (22 cm / 100 cm/m) / 65 cm/s
= 0.22 m / 65 s
≈ 0.00338 s
Now we can calculate the angular frequency:
angular frequency = 2π / period
= 2π / 0.00338 s
≈ 1860.96 rad/s
Therefore, the angular frequency of the particle is approximately 1860.96 rad/s.
b) To find the maximum force on the particle, we can use the equation:
maximum force = mass * angular frequency^2 * amplitude
Given:
mass (m) = 300 g = 0.3 kg
angular frequency (ω) = 1860.96 rad/s
amplitude (A) = 11 cm = 0.11 m (midpoint of the distance between the two extreme points)
Now we can calculate the maximum force:
maximum force = 0.3 kg * (1860.96 rad/s)^2 * 0.11 m
≈ 11436.35 N
Therefore, the maximum force on the particle is approximately 11436.35 N.
c) To find the maximum speed, we can use the equation:
maximum speed = angular frequency * amplitude
Given:
angular frequency (ω) = 1860.96 rad/s
amplitude (A) = 11 cm = 0.11 m (midpoint of the distance between the two extreme points)
Now we can calculate the maximum speed:
maximum speed = 1860.96 rad/s * 0.11 m
≈ 204.71 m/s
Therefore, the maximum speed of the particle is approximately 204.71 m/s.
To find the answers to these questions, we need to use the formulas related to Simple Harmonic Motion (SHM).
a) To find the angular frequency (ω) for a particle in SHM, we can use the formula:
ω = 2πf
where f is the frequency of oscillation. However, we are not given the frequency directly.
Using the given information, we know that the particle travels 11 cm between the two extreme points in its motion. This distance traveled is the amplitude (A) of the oscillation. We can find the frequency (f) using the formula:
f = v / λ
where v is the average speed of the particle and λ is the wavelength, which is equal to 2 times the amplitude.
Substituting the values into the formula, we have:
λ = 2A = 2 * 11 cm = 22 cm
f = 65 cm/s / 22 cm = 2.9545 Hz
Finally, we can find the angular frequency:
ω = 2π * f = 2π * 2.9545 Hz ≈ 18.556 rad/s
b) The maximum force on a particle in SHM can be found using the formula:
Fmax = mω²A
where m is the mass of the particle, ω is the angular frequency, and A is the amplitude.
Substituting the given values, we have:
Fmax = (0.300 kg) * (18.556 rad/s)² * (0.11 m) = 0.365 N
So, the maximum force on the particle is approximately 0.365 N.
c) The maximum speed (vmax) of the particle in SHM can be found using the formula:
vmax = ωA
Substituting the given values, we have:
vmax = (18.556 rad/s) * (0.11 m) = 2.04216 m/s
So, the maximum speed of the particle is approximately 2.04216 m/s.
a. period=11/65 seconds
angular freq=2PI/period=w
b. force=mass*acceleration
I assume you know a little calculus. Amplitude is 11/2
displacement=amplitude*sin(ang freq *time)
taking the derivative, w=angfreq
velocity= wA*cos(wt)
taking the second derivative
acceleration=-Aw^2sin(wt)
max acceleration=-A w^2
max force= mass*maxacceleration
max velocity (above) when cosine is 1, so vmax=wA