A mass of 0.440 kg is attached to a spring and set into oscillation on a horizon

Question

A mass of 0.440 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) (0.640 m)cos[ (16.0 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass (b) force constant for the spring N/m (c) position of the mass after it has been oscillating for one half a period (d) position of the mass one-sixth of a period after it has been released (e) time it takes the mass to get to the position x =-0.500 m after it has been released

Explanation / Answer

General equation for dispalcement of SHM as x(t) = A cos t

Here this is given as x(t) = 0.64 m cos (16t)

Comparing above two

Part A:

we have A = Amplitude = 0.64 m

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Part B:

Force Constant K = mw^2

K = 0.44* 16^2

K = 112.64 N/m

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Part C:

Time period T = 2pi/

T = (2*3.14)/(16)

T = 0.392 secs

At time t = T/2

t= 2pi/2 = pi/   

t=0.098 secs

So Position x(0.098) = 0.64 * cos (*pi/)

X(0.098) = -0.63999 m

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Part D:

at time t = T/3

t= 2pi/3

x(t) = 0.64m * cos (*2pi/3)

x(t) =-0.319 m       

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part E:

time taken to reach x = -0.5 m is

-0.5 = 0.64 cos (16 *t)

Cos (16 t) = -0.5/0.64 =-0.78125

16 t = 2.46

t = 0.154 secs

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