A body weighing 150 N, moves with simple harmonic motion. The velocity and
acceleration of the body when it is 200 mm from the centre of oscillation, are 5 m/s
and 20 m/s2 respectively.Determine (a) amplitude of motion ,no. of vibrations per minute, Periodic time and angular velocity
mass = 150/g or about 15 kg but who needs it?
x = A sin w t
v = Aw cos wt
a = -Aw^2 sin wt = -w^2 x
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when
x = .2 = A sin w t
v = 5 = Aw cos w t
a = +20=+Aw^2 sin w t = +w^2x
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so
w^2 x = w^2(.2) = 20
w^2 = 200/2 = 100
w = 10 radians/sec
now rewrite
.2 = A sin 10 t
5 = 10 A cos 10t
20 = 100A sin 10t
.2 = A sin 10 t
.5 = A cos 10 t
.04 = A^2 sin^2 10 t
.25 = A^2 cos^2 10 t
.04/A^2 + .25/A^2 = sin^2+cos^2 = 1
A^2 = .29
I guess you can take it from there
remember w = 2 pi f = 10
so f = 5/pi
To determine the amplitude of motion, we need to find the maximum displacement of the body from its equilibrium position.
Given:
Force (F) = 150 N
Acceleration (a) = 20 m/s^2
Velocity (v) = 5 m/s
Distance from centre of oscillation (x) = 200 mm = 0.2 m
To find the amplitude (A), we can use the equation:
F = m * a
where m is the mass of the body.
Since weight (W) = m * g, where g is the acceleration due to gravity, we can rewrite the equation as:
W = m * a
Therefore,
m * g = m * a
150 N = m * 20 m/s^2
Dividing both sides of the equation by g, we get:
m = 150 N / 20 m/s^2 = 7.5 kg
Now, to find the amplitude (A), we can use the equation:
A = x + v^2 / (a * x)
A = 0.2 m + (5 m/s)^2 / (20 m/s^2 * 0.2 m)
A = 0.2 m + 25 m^2/s^2 / 4 m^2/s^2
A = 0.2 m + 6.25 m
A = 6.45 m (approximately)
The amplitude of motion is approximately 6.45 meters.
To find the number of vibrations per minute, we can use the equation:
vibrations per minute = (1 / periodic time) * 60
The periodic time (T) is the time taken for one complete vibration.
To find T, we can use the equation:
T = 2π * √(m / k)
where k is the spring constant.
Since F = k * A, we can rewrite the equation as:
k = F / A
k = 150 N / 6.45 m
k = 23.26 N/m (approximately)
Now, substituting the values in the equation for T, we get:
T = 2π * √(7.5 kg / 23.26 N/m)
T = 2π * √(0.323 s^2)
T = 2π * 0.569 s
T = 3.58 s (approximately)
Now, substituting the value of T in the equation for vibrations per minute:
vibrations per minute = (1 / 3.58 s) * 60
vibrations per minute = 16.76 (approximately)
Therefore, the number of vibrations per minute is approximately 16.76.
Finally, to find the angular velocity (ω), we can use the equation:
ω = 2π / T
ω = 2π / 3.58 s
ω = 1.76 rad/s (approximately)
Therefore, the angular velocity is approximately 1.76 radians per second.
To determine the amplitude of motion, we can use the relationship between velocity and amplitude in simple harmonic motion. The velocity of the body is given as 5 m/s when it is 200 mm away from the center of oscillation.
We can use the equation v = ω√(A^2 - x^2), where v is the velocity, ω is the angular velocity, A is the amplitude, and x is the displacement from the center of oscillation. Rearranging the equation, we get A^2 = (v/ω)^2 + x^2.
Substituting the given values, we have:
5 = ω√((A^2) - (0.2^2))
Now, let's find the acceleration of the body when it is 200 mm from the center of oscillation. The acceleration is given as 20 m/s².
Using the equation a = ω^2√(A^2 - x^2), where a is the acceleration, ω is the angular velocity, A is the amplitude, and x is the displacement, we can substitute the values as follows:
20 = ω^2√((A^2) - (0.2^2))
Now, we have two equations:
5 = ω√((A^2) - (0.2^2))
20 = ω^2√((A^2) - (0.2^2))
We can solve these two equations simultaneously to find the amplitude and angular velocity.
Once we determine the amplitude and angular velocity, we can use them to calculate the number of vibrations per minute, periodic time, and angular velocity.
To calculate the number of vibrations per minute, we can use the relationship between frequency and periodic time. Frequency (f) is the reciprocal of the periodic time (T), and the number of vibrations per minute is f multiplied by 60:
Number of vibrations per minute = f * 60
To calculate the periodic time, we can use the equation:
T = 2π / ω
To calculate the angular velocity, we need to find the value of ω that satisfies both equations above. We can rearrange either equation to solve for ω and then substitute it into the other equation to find the amplitude and angular velocity.
Solving the equations and substituting the values will give us the answers for the amplitude of motion, the number of vibrations per minute, the periodic time, and the angular velocity.