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An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 1.6 cos(3.5t - 1.4), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.
a) Find the amplitude of the simple harmonic motion, in meters.
b) What is the value of the angular frequency, in radians per second?
c) Determine the position of the object, in meters, at the time t = 0.
d) What is the object s velocity, in meters per second, at time t = 0?
e) Calculate the object s acceleration, in meters per second squared, at time t = 0.
f) What is the magnitude of the object s maximum acceleration, in meters per second squared?

Sagot :

The amplitude of the object is 1.6m, the angular speed, position, velocity and acceleration of the object at t = 0 ate 3.5rad/s,  0.256m, 4.459m/s and 3.136m/s² and the maximum acceleration is 19.6m/s².

The object is undergoing Simple Harmonic Motion along x-axis and its position is described by,

x(t) = 1.6cos(3.5t-1.4)

t is the time.

(a) The standard equation of simple harmonic motion is,

X = Acos(wt)

So, comparing the given equation with the standard equation, we get,

Amplitude = 1.6m.

(b) Again comparing the given equation with standard equation, we get the angular velocity as,

W = 3.5rad/s.

(c) To find the position of the object at t=0,

Putting t=0 in x(t) = 1.6cos(3.5t-1.4).

x = 1.6cos(-1.4)

x = 1.6 × 0.16

x = 0.256m.

(d) To find the speed of the object,

Differentiating x(t) = 1.6cos(3.5t-1.4) with respect to x,

v(t) = 1.6sin(3.5t-1.4)3.5

Putting t = 0

v(t) = 1.6 x -0.98 x 3.5

v(t) = 4.459m/s.

(e) To find the acceleration of the object,

Double Differentiating x(t) = 1.6cos(3.5t-1.4) with respect to x,

a(t) = -1.6(3.5)²cos(3.5t-1.4)

Putting t = 0,

a(t) = -1.6 x (3.5)² x 0.16

a(t) = 3.136m/s².

(f) For maximum acceleration,

cos(3.5t-1.4) = 1

a(t) = -19.6m/s².

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