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Question: A particle executes simple harmonic motion with an amplitude of 10cm and time period 6s. At t=0 , it...

A particle executes simple harmonic motion with an amplitude of 10cm and time period 6s. At t=0 , it is at a distance if it is at point x=5cmx=5cm going towards positive xx direction.Write the equation of displacement (x) at time t . Find the magnitude of the acceleration of the particle at t =4s.

Explanation

Solution

Hint The equation of displacement (along x axis is given by: x=Acos⁡(wt+θ)x=A\cos \left( wt+\theta \right)) By putting initial condition values, the value of θ\theta and hence the required equation is achieved. Velocity can then be calculated by differentiating x with respect to t and hence, acceleration can be calculated by differentiating vv with respect tott .

Complete step by step solution
Equation of displacement along x axis is given by:
x=Acos⁡(cot⁡+θ) T=6 sec (given) ∴x=10 cos(2πT+t+θ) x=10cos(2π6+t+θ)  At t=0,x=5cm(given) \begin{aligned} & x=A\cos \left( \cot +\theta \right) \\\ & T=6\text{ sec (given)} \\\ & \therefore x=10\text{ cos}\left( \dfrac{2\pi }{T}+t+\theta \right) \\\ & x=10\text{cos}\left( \dfrac{2\pi }{6}+t+\theta \right) \\\ & \text{ At }t=0,x=5cm(\text{given)} \\\ \end{aligned}
∴5=10cos(π3×0+θ) cos⁡θ=12 θ−π3 ∴Equation of displacement along x axis x=10cos(2π6t+π3) \begin{aligned} & \therefore \text{5=}10\text{cos}\left( \dfrac{\pi }{3}\times 0+\theta \right) \\\ & \cos \theta =\dfrac{1}{2} \\\ & \theta -\dfrac{\pi }{3} \\\ & \therefore \text{Equation of displacement along }x\ \text{axis} \\\ & x=10\text{cos}\left( \dfrac{2\pi }{6}t+\dfrac{\pi }{3} \right) \\\ \end{aligned}
Now , we have to calculate velocity
It is given by:
v=dydx=−10sin⁡(2π6t+π3)×π3 acceleration a=dydx=−10(π3)2cos⁡(2π6t+π3) acceleration at t=4 \begin{aligned} & v=\dfrac{dy}{dx}=-10\sin \left( \dfrac{2\pi }{6}t+\dfrac{\pi }{3} \right)\times \dfrac{\pi }{3} \\\ & \text{acceleration }a=\dfrac{dy}{dx}=-10{{\left( \dfrac{\pi }{3} \right)}^{2}}\cos \left( \dfrac{2\pi }{6}t+\dfrac{\pi }{3} \right) \\\ & \text{acceleration at }t=4 \\\ \end{aligned}
 a=−10(π3)2cos⁡(2π6×4+π3)  =10(π3)2cos⁡(300∘)  =10(π3)2cos⁡(12)  a=−5⋅48cm/sec⁡2 ∴magnitude of acceleration is given by  a=−5⋅48cm/sec⁡2   \begin{aligned} & \text{ }a=-10{{\left( \dfrac{\pi }{3} \right)}^{2}}\cos \left( \dfrac{2\pi }{6}\times 4+\dfrac{\pi }{3} \right) \\\ & \text{ }=10{{\left( \dfrac{\pi }{3} \right)}^{2}}\cos \left( 300{}^\circ \right) \\\ & \text{ }=10{{\left( \dfrac{\pi }{3} \right)}^{2}}\cos \left( \dfrac{1}{2} \right) \\\ & \text{ }a=-5\cdot 48cm/{{\sec }^{2}} \\\ & \therefore \text{magnitude of acceleration is given by} \\\ & \text{ }a=-5\cdot 48cm/{{\sec }^{2}} \\\ & \text{ } \\\ \end{aligned}

Note Note that when displacement along x axis is given, the formula to be used isx=Acos⁡(wt+θ)x=A\cos \left( wt+\theta \right)
Where displacement along y axis is given, then formula to be used is y=Asin⁡(wt+θ)y=A\sin \left( wt+\theta \right)