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
To determine the amplitude of motion, we use the equation for velocity in simple harmonic motion:
v = ωA
where v is the velocity, ω is the angular velocity, and A is the amplitude.
Given that the velocity is 5 m/s, we can rearrange the equation to solve for A:
A = v / ω
To determine the angular velocity, we use the equation for acceleration in simple harmonic motion:
a = ω^2A
Given that the acceleration is 20 m/s^2 and the amplitude is A, we can rearrange the equation to solve for ω:
ω = √(a / A)
Now, let's substitute the given values into the equations:
A = 5 m/s / ω
ω = √(20 m/s^2 / A)
To find the amplitude, we need to find ω. So, let's substitute the ω equation into the amplitude equation:
A = 5 m/s / √(20 m/s^2 / A)
Squaring both sides of the equation, we get:
A^2 = (5 m/s)^2 / (20 m/s^2 / A)
Simplifying further:
A^2 = 25 m^2/s^2 / (20 m/s^2 / A)
A^2 = (25 m^2/s^2 * A) / (20 m/s^2)
A^2 = 1.25 m^2
Taking the square root of both sides, we find:
A ≈ 1.118 m
Therefore, the amplitude of motion is approximately 1.118 m.
To find the number of vibrations per minute, we need to determine the periodic time, which is the time taken for one complete vibration or oscillation.
The equation for the periodic time is:
T = 2π / ω
Given that ω = 2πf, where f is the frequency, we can substitute this into the equation to solve for T:
T = 2π / (2πf)
Simplifying:
T = 1 / f
To find the frequency, we can use the equation:
f = n / T
where n is the number of vibrations (or cycles) in a given time and T is the periodic time.
In this case, since we need the number of vibrations per minute and the periodic time in seconds, we can use n = 1 and T = 60 seconds:
f = 1 / (60 s) = 0.0167 Hz
Therefore, the frequency is approximately 0.0167 Hz.
Now, to find the periodic time, we use the equation:
T = 1 / f = 1 / 0.0167 Hz = 60 seconds
Therefore, the periodic time is 60 seconds.
Finally, to find the angular velocity ω, we substitute the frequency f into the equation ω = 2πf:
ω = 2π * 0.0167 Hz = 0.105 radians/second
Therefore, the angular velocity is approximately 0.105 radians/second.
To determine the amplitude of the motion, we'll use the equation for velocity in simple harmonic motion:
v = ω * A
Where v is the velocity, ω is the angular velocity, and A is the amplitude.
Given that the velocity v is 5 m/s and the body is 200 mm (0.2 m) from the center of oscillation, we can substitute these values into the equation:
5 = ω * 0.2
Solving for ω, we have:
ω = 5 / 0.2 = 25 rad/s
Now, let's determine the number of vibrations per minute. The number of vibrations per minute can be calculated using the formula:
Number of vibrations per minute = 60 / T
Where T is the periodic time.
The periodic time (T) can be calculated using the formula:
T = 2π / ω
Substituting the value of ω we calculated earlier:
T = 2π / 25 = 0.2513 s (approximately)
Now, let's calculate the number of vibrations per minute:
Number of vibrations per minute = 60 / 0.2513 = 238.4 (approximately)
So, the amplitude of motion is 0.2 m, the number of vibrations per minute is approximately 238.4, the periodic time is approximately 0.2513 s, and the angular velocity is 25 rad/s.