Question

X1= 3.75 sin (100πt + 2π/9)

x2 = 4.42 sin (100πt – 2π/5 )
At what time does each vibration first reach a displacement of 2mm?

Answer 1

For
`\(x_1\), \(t \approx 0.062\)`seconds. For
`\(x_2\), \(t \approx 0.077\)` seconds, when each vibration first reaches a 2mm displacement.

To find the time at which each vibration first reaches a displacement of 2mm, we can set the displacement equations equal to 2mm and solve for time. The displacement equations for
`\(x_1\) and \(x_2\)` are given by:

`\[ x_1 = 3.75 \sin(100\pi t + (2\pi)/(9)) \]`

`\[ x_2 = 4.42 \sin(100\pi t - (2\pi)/(5)) \]`

Setting each equation equal to 2mm, we get:

`\[ 3.75 \sin(100\pi t + (2\pi)/(9)) = 2 \]`

`\[ 4.42 \sin(100\pi t - (2\pi)/(5)) = 2 \]`

To find the time t, we need to solve these equations. Due to the periodic nature of sine functions, there will be multiple solutions. The general solution to the equation
`\(a \sin(\theta) = b\) is \(\theta = \sin^(-1)((b)/(a)) + 2\pi n\)`, where n is an integer.

Once you find the solutions for both equations, you can choose the smallest positive values of t since you are looking for the time when each vibration first reaches a displacement of 2mm.

Keep in mind that the time t should be in seconds, considering the angular frequency
`\(100\pi\)` in the equations.