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hboard My courses A point in SHM (Simple harmonic motion) has an acceleration of \( 4.56 \mathrm{~m} / \mathrm{s}^{2} \) at a distance 0.4 mm from the origin. Note: Please return your results up to the 4 th decimal point e.g 3.2344 given the sensitivity of results in the problem After analysis of the defined SHM, the period of oscillations is \( \boldsymbol{T}_{\boldsymbol{p}}= \) \( \square \) \( s \) (1) and The frequency is \( f= \) \( \square \) Hz Assuming that the Amplitude of motion is \( \mathbf{0 . 4 8 m m} \) with an intitial phase of \( \frac{\boldsymbol{\pi}}{\mathbf{4}} \mathbf{r a d} \) develop the applicable equation of motion \( x=x_{\text {max }} \cos \left(\omega t+\phi_{o}\right) \) and calculate the position of the point at at the initial instant \( t_{o}=0 s: x\left(t_{o}\right)= \) \( \square \) \( \boldsymbol{m} \boldsymbol{m} \) and then calculate: \[ x(0.04 s)=\quad m m \quad x(0.89 s)=\quad m m \] \[ v(0.04 s)=\quad \mathrm{m} / \mathrm{s} \] \( \qquad \) \[ a(0.04 s)=\quad m / s^{2} \quad a(0.89 s)=\quad m / s^{2} \] \( ? \) The time \( \Delta t \) required for a point in SHM to move from the center of motion to the Amplitude is \( \square \) of the Search

          hboard My courses

A point in SHM (Simple harmonic motion) has an acceleration of \( 4.56 \mathrm{~m} / \mathrm{s}^{2} \) at a distance 0.4 mm from the origin.
Note: Please return your results up to the 4 th decimal point e.g 3.2344 given the sensitivity of results in the problem
After analysis of the defined SHM, the period of oscillations is \( \boldsymbol{T}_{\boldsymbol{p}}= \) \( \square \) \( s \) (1) and

The frequency is \( f= \) \( \square \) Hz

Assuming that the Amplitude of motion is \( \mathbf{0 . 4 8 m m} \) with an intitial phase of \( \frac{\boldsymbol{\pi}}{\mathbf{4}} \mathbf{r a d} \) develop the applicable equation of motion \( x=x_{\text {max }} \cos \left(\omega t+\phi_{o}\right) \) and calculate the position of the point at at the initial instant \( t_{o}=0 s: x\left(t_{o}\right)= \)
\( \square \) \( \boldsymbol{m} \boldsymbol{m} \) and then calculate:
\[
x(0.04 s)=\quad m m \quad x(0.89 s)=\quad m m
\]
\[
v(0.04 s)=\quad \mathrm{m} / \mathrm{s}
\]
\( \qquad \)
\[
a(0.04 s)=\quad m / s^{2} \quad a(0.89 s)=\quad m / s^{2}
\]
\( ? \)

The time \( \Delta t \) required for a point in SHM to move from the center of motion to the Amplitude is \( \square \) of the
Search

hboard My courses

A point in SHM (Simple harmonic motion) has an acceleration of 4.56 m / s^2 at a distance 0.4 mm from the origin.
Note: Please return your results up to the 4 th decimal point e.g 3.2344 given the sensitivity of results in the problem
After analysis of the defined SHM, the period of oscillations is Tp= □ s (1) and

The frequency is f= □ Hz

Assuming that the Amplitude of motion is 0 . 4 8 𝐦 𝐦 with an intitial phase of (π)/(4)𝐫 𝐚 𝐝 develop the applicable equation of motion x=xmax cos(ω t+ϕo) and calculate the position of the point at at the initial instant to=0 s: x(to)=
□ mm and then calculate:

x(0.04 s)= m m x(0.89 s)= m m

v(0.04 s)= m / s

a(0.04 s)= m / s^2 a(0.89 s)= m / s^2

?

The time Δ t required for a point in SHM to move from the center of motion to the Amplitude is □ of the
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Added by Darnell C.

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