Answer:
x(t) = 0.053 sin(8t) + 0.425t + 0.1
Explanation:
Using Newton’s 2nd law we can calculate the acceleration equation of the block with respect to time t
[tex]a(t) = F/m = 17 sin(8t) / 5 = 3.4 sin(8t)[/tex]
From here we can derive the velocity equation by integrating the acceleration equation
[tex]v(t) = \int\limits {a(t)} \, dt =- 3.4 cos(8t)/8 + v_0 = -0.425 cos(8t) + C_v[/tex]
Since the block is released from rest at t = 0, it means the velocity is 0 at t = 0
[tex]v(0) = 0[/tex]
[tex]-0.425cos(0) + C_v = 0[/tex]
[tex]C_v = 0.425 m/s[/tex]
[tex]v(t) = -0.425 cos(8t) + 0.425 [/tex]
To find the equation of motion, we integrate the velocity equation:
[tex]x(t) = \int\limit {v(t)} \, dt = -0.425sin(8t)/8 + 0.425t + C_x[/tex]
[tex]x(t) = 0.053 sin(8t) + 0.425t + C_x[/tex]
At t = 0, object is released at position x = 100 mm = 0.1 m:
[tex]x(0) = 0.1[/tex]
[tex] 0.053 sin(0) + 0.425*0 + C_x = 0.1[/tex]
[tex]C_x = 0.1 [/tex]
[tex]x(t) = 0.053 sin(8t) + 0.425t + 0.1[/tex]
