Question

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Answer 1

GiVeN :-

The displacement of a moving particle is represented as,

$\bf{\red{\longrightarrow}}\:x\:=\:4\:\sin(4t)\:+\:3\:\cos(4t)\:$

x is in centimetre .

t is in second .

To FiNd :-

$\bf{\pink{(1)}}$ It's motion in S.H.M .

$\bf{\pink{(2)}}$ It's time period .

$\bf{\pink{(3)}}$ It's maximum velocity .

$\bf{\pink{(4)}}$ It's maximum acceleration .

SoLuTiOn :-

◍ To convert the given displacement equation in it's motion in S.H.M, we can find that equation as in the form of '$\bf{x\:=\:A\:\sin\:(\omega{t}\:+\:\phi)\:}$' .

$\bf{\red{\longrightarrow}}\:x\:=\:4\:\sin(4t)\:+\:3\:\cos(4t)\:$

$\bf{\red{\longrightarrow}}\:\dfrac{x}{5}\:=\:\dfrac{4}{5}\:.\:\sin(4t)\:+\:\dfrac{3}{5}\:.\:\cos(4t)\:$

Take as,

$\bf{\dfrac{4}{5}\:=\:\cos{\alpha}}$

And $\bf{\dfrac{3}{5}\:=\:\sin{\alpha}}$

$\bf{\red{\longrightarrow}}\:\dfrac{x}{5}\:=\:\cos{\alpha}\:.\:\sin(4t)\:+\:\sin{\alpha}\:.\:\cos(4t)\:$

$\bf{\red{\longrightarrow}}\:\dfrac{x}{5}\:=\:\sin\:(4t\:+\:\alpha)\:$

$\bf{\red{\longrightarrow}}\blue{\:x\:=\:5\:\sin\:(4t\:+\:\alpha)\:}$

➤ Comparing with the standard equation, we get

A = amplitude = 5

$\bf{\omega}$ = 4

$\bf{\phi}$ = $\bf{\alpha}$ = $\bf{\cos^{-1}{\Big(\dfrac{4}{5}\Big)}}$ = $\bf{\sin^{-1}{\Big(\dfrac{3}{5}\Big)}}$

∴ The given displacement equation is the motion of S.H.M .

___________________________

We know that,

$\huge\star$ $\bf\orange{Time\:period\:=\:\dfrac{2\:\pi}{\omega}\:}$

$\rm{\implies\:T\:=\:\dfrac{2\:\pi}{4}\:}$

$\bf{\implies\:T\:=\:\dfrac{\pi}{2}\:sec\:}$

∴ It's time period is $\bf{\dfrac{\pi}{2}\:sec\:}$ .

___________________________

We know that,

$\huge\star$ $\bf\orange{v_{max}\:=\:A\:\omega\:}$

$\rm{\implies\:v_{max}\:=\:5\times{4}\:}$

$\bf{\implies\:v_{max}\:=\:20\:cm.s^{-1}\:}$

∴ It's maximum velocity is $\bf{20\:cm.s^{-1}\:}$ .

___________________________

We know that,

$\huge\star$ $\bf\orange{a_{max}\:=\:A\:{\omega}^2\:}$

$\rm{\implies\:a_{max}\:=\:5\times{4^2}\:}$

$\bf{\implies\:a_{max}\:=\:80\:cm.s^{-2}\:}$

∴ It's maximum acceleration is $\bf{80\:cm.s^{-2}\:}$ .

Answer 2

Answer :-

To convert the given displacement equation in it's motion in S.H.M, we can find that equation as in the form of '$\bf{x\:=\:A\:\sin\:(\omega{t}\:+\:\phi)\:}$

$\begin{gathered}\bf{\red{\longrightarrow}}\:x\:=\:4\:\sin(4t)\:+\:3\:\cos(4t)\: \\ \end{gathered}$

$\begin{gathered}\bf{\red{\longrightarrow}}\:\dfrac{x}{5}\:=\:\dfrac{4}{5}\:.\:\sin(4t)\:+\:\dfrac{3}{5}\:.\:\cos(4t)\: \\ \end{gathered}$

Take as,

$\bf{\dfrac{4}{5}\:=\:\cos{\alpha}}$

And $\begin{gathered}\bf{\dfrac{3}{5}\:=\:\sin{\alpha}} \\ \end{gathered}$

$\begin{gathered}\bf{\red{\longrightarrow}}\:\dfrac{x}{5}\:=\:\cos{\alpha}\:.\:\sin(4t)\:+\:\sin{\alpha}\:.\:\cos(4t)\: \\ \end{gathered}$

$\begin{gathered}\bf{\red{\longrightarrow}}\:\dfrac{x}{5}\:=\:\sin\:(4t\:+\:\alpha)\: \\ \end{gathered}$

$\begin{gathered}\bf{\red{\longrightarrow}}\blue{\:x\:=\:5\:\sin\:(4t\:+\:\alpha)\:} \\ \end{gathered}$

➤ Comparing with the standard equation, we get

A = amplitude = 5

$\bf{\omega}ω = 4$

$\bf{\phi}ϕ = \bf{\alpha}α = \bf{\cos^{-1}{\Big(\dfrac{4}{5}\Big)}}$

$) = \begin{gathered}\bf{\sin^{-1}{\Big(\dfrac{3}{5}\Big)}} \\ \end{gathered}$

∴ The given displacement equation is the motion of S.H.M .

___________________________

We know that,

$\huge\star⋆ \begin{gathered}\bf\orange{Time\:period\:=\:\dfrac{2\:\pi}{\omega}\:}$

$\begin{gathered}\rm{\implies\:T\:=\:\dfrac{2\:\pi}{4}\:} \\ \end{gathered}$

$\begin{gathered}\bf{\implies\:T\:=\:\dfrac{\pi}{2}\:sec\:} \\ \end{gathered}$

∴ It's time period is $\bf{\dfrac{\pi}{2}\:sec\:}$

___________________________

We know that,

$\huge\star⋆ \begin{gathered}\bf\orange{v_{max}\:=\:A\:\omega\:} \\ \end{gathered}$

$\begin{gathered}\rm{\implies\:v_{max}\:=\:5\times{4}\:} \\ \end{gathered}$

$\begin{gathered}\bf{\implies\:v_{max}\:=\:20\:cm.s^{-1}\:} \\ \end{gathered}$

∴ It's maximum velocity is $\bf{20\:cm.s^{-1}\:}$20cm.

___________________________

We know that,

$\huge\star$ $\begin{gathered}\bf\orange{a_{max}\:=\:A\:{\omega}^2\:} \\ \end{gathered}$

$\begin{gathered}\rm{\implies\:a_{max}\:=\:5\times{4^2}\:} \\ \end{gathered}$

$\begin{gathered}\bf{\implies\:a_{max}\:=\:80\:cm.s^{-2}\:} \\ \end{gathered}$

∴ It's maximum acceleration is $\bf{80\:cm.s^{-2}\:}$80cm.