Question

The displacement of moving particle is x=4sin(4t)+3cos(4t) where x is in cm and t is in seconds.Show that its motion in SHM , calculate its time period and maximum velocity and maximum acceleration
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Answer 1

Given, the displacement of the particle,

$\sf{\longrightarrow x=4\sin(4t)+3\cos(4t)}$

Dividing both sides by 5,

$\sf{\longrightarrow\dfrac{x}{5}=\dfrac{4}{5}\sin(4t)+\dfrac{3}{5}\cos(4t)}$

Let $\alpha$ be an angle such that $\sf{\cos\alpha=\dfrac{4}{5}}$ and $\sf{\sin\alpha=\dfrac{3}{5}.}$

Then,

$\sf{\longrightarrow\dfrac{x}{5}=\sin(4t)\cos\alpha+\cos(4t)\sin\alpha}$

$\sf{\longrightarrow\dfrac{x}{5}=\sin(4t+\alpha)\quad\quad\![\,\sin(a+b)=\sin a\cos b+\cos a\sin b\,]}$

$\sf{\longrightarrow x=5\sin(4t+\alpha)}$

Now the equation is in the form $\sf{x=A\sin(\omega t+\phi)}$ where,

• $\sf{A=5\ cm\quad\quad\![\,x\ is\ in\ cm\,]}$

• $\sf{\omega=4\ rad\,s^{-1}}$

• $\sf{\phi=\alpha=\cos^{-1}\left(\dfrac{4}{5}\right)=\sin^{-1}\left(\dfrac{3}{5}\right)}$

Hence the motion of the particle is SHM.

Time period of the particle,

$\sf{\longrightarrow T=\dfrac{2\pi}{\omega}}$

$\sf{\longrightarrow T=\dfrac{2\pi\ rad}{4\ rad\,s^{-1}}}$

$\sf{\longrightarrow\underline{\underline{T=\dfrac{\pi}{2}\ s}}}$

Maximum velocity of the particle,

$\sf{\longrightarrow v_{max}=A\omega}$

$\sf{\longrightarrow v_{max}=5\ cm\times4\ rad\,s^{-1}}$

$\sf{\longrightarrow\underline{\underline{v_{max}=20\ cm\,s^{-1}}}}$

Maximum acceleration of the particle,

$\sf{\longrightarrow a_{max}=A\omega^2}$

$\sf{\longrightarrow a_{max}=5\ cm\times4^2\ rad^2\,s^{-2}}$

$\sf{\longrightarrow\underline{\underline{a_{max}=80\ cm\,s^{-2}}}}$

Answer 2

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 $\bf{80\:cm.s^{-2}\:}$ .