A particle starts from a point $P$ at a distance of $A / 2$ from the mean position $O$ & travels towards left as shown in the figure. If the time period of SHM, executed about $O$ is $T$ and amplitude $A$ then the equation of motion of particle is :

x = A sin (\frac{2 π}{T} t + \frac{π}{6})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = A sin (\frac{2 π}{T} t + \frac{5 π}{6})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = A cos (\frac{2 π}{T} t + \frac{π}{6})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = A cos (\frac{2 π}{T} t + \frac{π}{3})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x = A sin (\frac{2 π}{T} t + \frac{5 π}{6})$
$\begin{matrix} x = A cos (ω t + θ) a n d v = - ω A sin (ω t + θ) a t t = 0, x = A c o s θ a n d v = - ω A sin θ ∴ \frac{A}{2} = A c o s θ \Rightarrow θ = {cos}^{- 1} (\frac{1}{2}) = \frac{π}{3} n o w, x = A cos (ω t + \frac{π}{3}) = A cos (\frac{2 π}{T} . t + \frac{π}{3}) = A s i n (\frac{2 π}{T} . t + \frac{π}{2} + \frac{π}{3}) ∴ A s i n (\frac{2 π}{T} . t + \frac{5 π}{6}) s o t h e c o r r e c t o p t i o n i s B . \end{matrix}$

Suggest Corrections

Similar questions

Q. A particle starts from point

x = - \frac{\sqrt{3}}{2} A

and move towards negative extreme as shown in the figure below. If the time period of oscillation is

T

, then

Q. displacement time equation of a particle executing SHM is, x=sin[(π/3)t+π/6]cm. The distance of a particle in 3 seconds is

The equation of a particle executing SHM is $x = (5 m) s i n [(π s^{- 1}) t + \frac{π}{3}]$ . What is its amplitude, time period, maximum speed and velocity. At t = 1s?

Q. A particle performing SHM with amplitude

A

and time period

T

starts from mean position towards the positive extreme point as shown in the figure. Find the time taken by the particle (in seconds) to reach

x = \frac{A}{2} .

Q. The equation of a particle executing SHM is

x = (5 m) sin [(π s^{- 1} t = \frac{π}{6}]

. Write down the amplitude, initial phase constant, time period and maximum speed.

Join BYJU'S Learning Program

A particle starts from a point P at a distance of A/2 from the mean position O & travels towards left as shown in the figure. If the time period of SHM, executed about O is T and amplitude A then the equation of motion of particle is :

The correct option is B x=Asin(2πTt+5π6)x=Acos(ωt+θ)andv=−ωAsin(ωt+θ)att=0,x=Acosθandv=−ωAsinθ∴A2=Acosθ⇒θ=cos−1(12)=π3now,x=Acos(ωt+π3)=Acos(2πT.t+π3)=Asin(2πT.t+π2+π3)∴Asin(2πT.t+5π6)sothecorrectoptionisB.

A particle starts from a point $P$ at a distance of $A / 2$ from the mean position $O$ & travels towards left as shown in the figure. If the time period of SHM, executed about $O$ is $T$ and amplitude $A$ then the equation of motion of particle is :