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
Plase solve this problem
Well, this problem seems to be oscillating between different units of measurement. Let me try solving it while trying not to get too tangled up in the confusion.
Let's start by converting everything to the same unit, shall we? So, 200 mm is equal to 0.2 meters.
Now, we can go ahead and determine the amplitude of the motion. The amplitude is the maximum displacement from the center of oscillation, which in this case is given as 0.2 meters.
To find the number of vibrations per minute, we need to know the periodic time. The periodic time is the time taken for one complete oscillation. We can find it using the formula:
T = 1/f
Where T is the periodic time and f is the frequency. The frequency can be found using the formula:
f = 1/T
Now, we need to find the angular velocity, which is given by the formula:
ω = 2πf
Where ω is the angular velocity and f is the frequency.
Let's plug in the values and see what we get, shall we?
T = 1/f
f = 1/T
Taking the reciprocal of T, we get:
T = 1/((200 mm)/5 m/s)
T = 1/((0.2 m)/5 m/s)
T = 1/(0.04 s)
T = 25 s
So, the periodic time is 25 seconds.
Now, let's find the frequency:
f = 1/T
f = 1/25 s
f = 0.04 Hz
So, the frequency is 0.04 Hz.
Now, let's find the angular velocity:
ω = 2πf
ω = 2π(0.04 Hz)
ω = 0.08π rad/s
So, the angular velocity is 0.08π rad/s.
Phew! I hope I didn't lose you in the unit conversion madness. But hey, at least we made it through in the end!
Given:
Weight of the body, W = 150 N
Velocity, v = 5 m/s
Acceleration, a = 20 m/s^2
Distance from center of oscillation, x = 200 mm = 0.2 m
To find:
(a) Amplitude of motion
(b) Number of vibrations per minute
(c) Periodic time
(d) Angular velocity
Solution:
(a) The amplitude of motion (A) can be calculated using the formula:
A = x / sinθ
Where, θ is the angle between the position of the body and the center of oscillation.
We can calculate the angle using the formula:
sinθ = v / √(ω^2 - (a/g)^2)
Given: a = 20 m/s^2, g = 9.8 m/s^2
Calculating the value of ω:
a/g = 20 / 9.8 = 2.04
Now, ω^2 = (v / x)^2 + (a / g)^2
ω^2 = (5 / 0.2)^2 + 2.04^2
ω^2 = 625 + 4.1616
ω^2 = 629.1616
Taking square root on both sides:
ω = √629.1616
ω ≈ 25.08 rad/s
Now substituting the value of ω in the equation of sinθ:
sinθ = v / √(ω^2 - (a / g)^2)
sinθ = 5 / √(629.1616 - 4.1616)
sinθ = 5 / √(625 + 4.1616)
sinθ = 5 / √629.1616
sinθ ≈ 0.126
Therefore, the amplitude of motion (A) can be calculated as:
A = x / sinθ
A = 0.2 / 0.126
A ≈ 1.5873 m
So, the amplitude of motion is approximately 1.5873 m.
(b) The number of vibrations per minute can be calculated using the formula:
Number of vibrations per minute = (60 * v) / λ
Where, v is the velocity and λ is the wavelength.
Wavelength can be calculated as:
λ = 2πA
Substituting the values of A and v:
λ = 2π(1.5873)
λ ≈ 9.9568 m
Therefore, the number of vibrations per minute can be calculated as:
Number of vibrations per minute = (60 * v) / λ
Number of vibrations per minute = (60 * 5) / 9.9568
Number of vibrations per minute ≈ 30.234
So, the number of vibrations per minute is approximately 30.234.
(c) The periodic time (T) can be calculated using the formula:
T = 1 / (v / A)
Substituting the given values:
T = 1 / (5 / 1.5873)
T = 1 / 3.1546
T ≈ 0.3172 s
Therefore, the periodic time is approximately 0.3172 seconds.
(d) The angular velocity (ω) can be calculated using the formula:
ω = 2π / T
Substituting the value of T:
ω = 2π / 0.3172
ω ≈ 19.794 rad/s
Therefore, the angular velocity is approximately 19.794 rad/s.
To solve this problem, let's break it down step-by-step and use the formulas for simple harmonic motion.
Step 1: Find the amplitude of motion.
The amplitude of motion is the maximum displacement from the equilibrium position. In this case, the body is 200 mm from the centre of oscillation.
Amplitude (A) = 200 mm = 200/1000 = 0.2 m
So, the amplitude of motion is 0.2 m.
Step 2: Find the number of vibrations per minute.
The number of vibrations per minute is a measure of frequency. It is given by the formula:
Number of vibrations per minute = 60/t
To find the value of t, we need to use the formula for the periodic time (T) in terms of frequency (f):
T = 1/f
Given that the body is in simple harmonic motion, we know that its acceleration (a) is related to the angular velocity (ω) and amplitude (A) by the formula:
a = ω^2A
Step 3: Find the periodic time (T) and angular velocity (ω).
From the given information, we know that the acceleration (a) is 20 m/s^2 and the amplitude (A) is 0.2 m.
Plugging these values into the formula, we can solve for the angular velocity (ω):
20 = ω^2 * 0.2
ω^2 = 20/0.2
ω^2 = 100
ω = sqrt(100)
ω = 10 rad/s
Now that we have the angular velocity (ω), we can use it to find the periodic time (T):
T = 2π/ω
T = 2π/10
T = π/5 seconds
Finally, we can find the number of vibrations per minute:
Number of vibrations per minute = 60/T
Number of vibrations per minute = 60/(π/5)
Number of vibrations per minute = 300/π
Number of vibrations per minute ≈ 95.49
So, the amplitude of motion is 0.2 m, the number of vibrations per minute is approximately 95.49, the periodic time is π/5 seconds, and the angular velocity is 10 rad/s.