The position of a 0.30 kg object attached to a spring is described by the expression ?(?) = (4.00 ?)cos(3.00?? + ?), where x is in meters and t is in seconds. Determine (a) the frequency and (b) period of the motion, (c) the amplitude of the motion, (d) the phase constant, (e) the spring constant, (f) the position of the object at t = 0.250 s, (g) the object’s velocity at t = 0.250 s, and (h) the object’s acceleration at t = 0.250 s.

The position of a 0.30 kg object attached to a spring is described by the expression ?(?) = (4.00 ?)cos(3.00?? + ?), where x is in meters and t is in seconds. Determine (a) the frequency and (b) period of the motion, (c) the amplitude of the motion, (d) the phase constant, (e) the spring constant, (f) the position of the object at t = 0.250 s, (g) the object’s velocity at t = 0.250 s, and (h) the object’s acceleration at t = 0.250 s.

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Question

The position of a 0.30 kg object attached to a spring is described by the expression
?(?) = (4.00 ?)cos(3.00?? + ?), where x is in meters and t is in seconds. Determine (a) the
frequency and (b) period of the motion, (c) the amplitude of the motion, (d) the phase
constant, (e) the spring constant, (f) the position of the object at t = 0.250 s, (g) the object’s
velocity at t = 0.250 s, and (h) the object’s acceleration at t = 0.250 s.

Expert Solution

Step 1

Given:

$x = 4.0 \cos (3.0 πt + π)$ ...(1)

From the below expression,

$x = A \cos (ω t + ϕ)$

The values of $ω = 3 π$

Time $T = \frac{2 π}{ω}$

$\Rightarrow T = \frac{2 π}{3 π}$

$\Rightarrow T = 0.67 s$

Step 2

a) The frequency is,

$F r e q u e n c y = \frac{1}{T i m e}$

$F r e q u e n c y = \frac{1}{0.67 s}$

$F r e q u e n c y = 1.5 H z$

Step by step

Solved in 3 steps

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