Find the amplitude, frequency, and period of motion for an object vibrating at t

Question

Find the amplitude, frequency, and period of motion for an object vibrating at the end of a horizontal spring if the equation for its position as a function of time is the following. x = (0.235 m) cos(pi/8.00 t) A = m f = Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Hz T = Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. s Find the maximum magnitude of the velocity and acceleration. V_max = Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s^2 a_max = Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s^2 What are the position, velocity, and acceleration of the object after 1.10 s has elapsed? X = Your response is within 10% of the correct value. This may be due to round off error, or you = could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize round off error. M v = The response you submitted has the wrong sign. m/s The response you submitted has the wrong sign. m/s a = Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s^2

Explanation / Answer

here,

(a)

x = (0.235 m) * cos(pi/8 * t )

amplitude , A = 0.235 m

w = pi/8 rad/s

2 * pi * f = pi/8

f = 0.0625 Hz

the frequency is 0.0625 Hz

time period , T= 1/f = 16 s

(b)

the maximum velocity , vm = A * w

vm = 0.092 m/s

the maximum accelration , am = A * w^2

am = 0.036 m/s^2

(c)

x = (0.235 m) * cos(pi/8 * t ) ...(1)

at t = 1.1 s

x = 0.213 m

differentiating equation (1)

v = - ( 0.092) * sin(pi/8 * t) ...(2)

at t = 1.1 s

v = - 0.038 m/s

differentiating equation (2)

a = - ( 0.036) * cos(pi/8 * t)

at t = 1.1 s

a = - 0.033 m/s^2

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