A body oscillates with simple harmonic motion along the x-axis. its displacement varies with time according to the equation x(t) = a sin(ω t + φ). if a = 5 m, ω = 3.444 rad/s, and φ = 1.0472 rad, what is the acceleration of the body at t = 3 s? note: the argument of the sine function is in radians rather than degrees. answer in units of m/s 2 .

The acceleration of the body at t = 3 seconds is approximately  [tex]\( -58.58785 \, \text{m/s}^2 \).[/tex]

To find the acceleration of the body at ( t = 3 ) seconds, we need to find the second derivative of the displacement function [tex]\( x(t) \)[/tex] with respect to time t.

Given that[tex]\( x(t) = a \sin(\omega t + \phi) \)[/tex] , where [tex]\( a = 5 \) m, \( \omega = 3.444 \) rad/s, and \( \phi = 1.0472 \)[/tex] rad, the acceleration a(t) is the second derivative of of x(t) with respect to t:

First, let's find the first derivative of of x(t) with respect to t:

[tex]\[ v(t) = \frac{dx}{dt} = a \omega \cos(\omega t + \phi) \][/tex]

Now, let's find the second derivative of x(t) with respect to t:

[tex]\[ a(t) = \frac{d^2x}{dt^2} = -a \omega^2 \sin(\omega t + \phi) \][/tex]

Now, substitute the given values:

[tex]\[ a(t) = -5 \times (3.444)^2 \sin(3.444 \times 3 + 1.0472) \][/tex]

Now, calculate:

[tex]\[ a(t) = -5 \times (3.444)^2 \sin(10.332 + 1.0472) \]\[ a(t) = -5 \times (3.444)^2 \sin(11.3792) \][/tex]

Now, compute:

[tex]\[ a(t) = -5 \times (11.858736) \sin(11.3792) \]\[ a(t) \approx -5 \times (11.858736) \times 0.97989 \]\[ a(t) \approx -58.58785 \, \text{m/s}^2 \][/tex]

Therefore, the acceleration of the body at t = 3 seconds is approximately  [tex]\( -58.58785 \, \text{m/s}^2 \).[/tex]

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