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

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To solve this problem, we can use the equations related to simple harmonic motion (SHM).

First, let's determine the amplitude of motion. The amplitude (A) is the maximum displacement from the mean position. In this case, the body is 200 mm (or 0.2 m) from the center of oscillation. Therefore, the amplitude is 0.2 m.

Next, let's find the number of vibrations per minute. The number of vibrations per minute is equal to the frequency of oscillation. The frequency (f) is the reciprocal of the periodic time (T), which is the time taken for one complete vibration.

To find the periodic time (T), we can use the equation: T = 1/f. Since we know the acceleration (a) and displacement (x), we can use the equation: a = (2πf)^2x.

Substituting the given values into the equation, we get: 20 m/s^2 = (2πf)^2 * 0.2 m.

To find the angular velocity (ω), we can use the equation: ω = 2πf.

Now, let's solve these equations to find the required values:

Step 1: Solving for f:
20 = (2πf)^2 * 0.2
20 = 4π^2f^2 * 0.2
f^2 = 20 / (4π^2 * 0.2)
f = √(20 / (4π^2 * 0.2))
f ≈ √(5 / (π^2 * 0.2))
f ≈ √(5 / (3.14^2 * 0.2))
f ≈ √(5 / 1.2552)
f ≈ √3.98
f ≈ 1.99 Hz (rounded to 2 decimal places)

Step 2: Solving for T:
T = 1/f
T ≈ 1 / 1.99
T ≈ 0.50 s (rounded to 2 decimal places)

Step 3: Solving for ω:
ω = 2πf
ω ≈ 2π * 1.99
ω ≈ 12.48 rad/s (rounded to 2 decimal places)

Therefore, the amplitude of motion is 0.2 m, the number of vibrations per minute is approximately 1.99 Hz, the periodic time is approximately 0.50 s, and the angular velocity is approximately 12.48 rad/s.