The displacement of a block attached to a horizontal spring is given by
x=0.5sin(7.8t−1.06)m
What is the earliest time (t>0) at which x=0.16m?
Find the acceleration when x=0.44m
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To find the earliest time (t>0) at which x=0.16m, we can set the given displacement equation equal to 0.16m and solve for t.
Given equation: x=0.5sin(7.8t−1.06)m
0.16 = 0.5sin(7.8t−1.06)
To solve this equation, we need to isolate sin(7.8t−1.06). First, divide both sides of the equation by 0.5:
0.16 / 0.5 = sin(7.8t−1.06)
0.32 = sin(7.8t−1.06)
Next, we need to find the inverse sine (also known as arcsine) of both sides of the equation to solve for the angle (7.8t−1.06). Since sin(θ) = 0.32, we have:
7.8t−1.06 = arcsin(0.32)
Now, to solve for (7.8t−1.06), we can add 1.06 to both sides:
7.8t = arcsin(0.32) + 1.06
Finally, divide both sides by 7.8 to isolate 't':
t = (arcsin(0.32) + 1.06) / 7.8
Using a calculator, evaluate the expression on the right-hand side to find the value of 't'. This will give you the earliest time at which x=0.16m.
To find the acceleration when x = 0.44m, we can differentiate the displacement equation with respect to time (t) to obtain the velocity equation. Then, we can differentiate the velocity equation with respect to time to obtain the acceleration equation.
Given equation: x = 0.5sin(7.8t−1.06)m
To find the velocity equation, differentiate x with respect to t:
v = dx/dt = 0.5 * d(sin(7.8t−1.06))/dt
Recall that the derivative of sine is cosine:
v = 0.5 * 7.8 * cos(7.8t−1.06)
Simplifying:
v = 3.9 * cos(7.8t−1.06)
Now, to find the acceleration equation, differentiate v with respect to t:
a = dv/dt = d(3.9 * cos(7.8t−1.06))/dt
Differentiating cosine gives us:
a = -3.9 * 7.8 * sin(7.8t−1.06)
Simplifying:
a = -30.42 * sin(7.8t−1.06)
Now, substitute x = 0.44m into the acceleration equation:
0.44 = -30.42 * sin(7.8t−1.06)
You can solve this equation numerically using a calculator or a computer program to find the value of 't' when x = 0.44m.