Question

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

Answer 1

Hint: The position of particle at any time can be expressed as a function of time. Initially, the position of the particle at the starting is found out. Then, the position of the particle at any time ‘t’ is found. From these, the magnitude of acceleration of the particle at any time ‘t’ is found.

Complete answer:
Let x be the position of the particle.
A be the amplitude
t be the time taken
ω is the angular velocity
And,
Φ be the phase angle
The position of a particle can be expressed as a function of time,
\[x\text{ }=\text{ }A\text{ }sin\text{ }\left( \omega t\text{ }+\text{ }\Phi \right)\]----(1)
Let this be equation (1)
In the above equation,
\[\omega t+\Phi \]is the phase of the particle
 The initial phase can be said as the phase angle at time t = 0
Substituting the values of x, A and t at x = 5 in equation (1)
we get
\[5=10\sin (\omega \text{x}0+\Phi )\]
\[5=10\sin (\Phi )\]----(2)
Let this be equation (2)
Dividing equation (2) by 10,
We get,
\[\dfrac{1}{2}=\sin (\Phi )\] ---(3)
Let this be equation (3)
From equation (3),
We get
\[\Phi ={{\sin }^{-1}}(\dfrac{1}{2})\]----(4)
Let this be equation (4)
Taking the sin inverse of
\[(\dfrac{1}{2})\]
We get,
\[\Phi =\dfrac{\pi }{6}\] ---(5)
Let this be equation (5)
We know that the angular velocity
\[\omega =\dfrac{2\pi }{T}\] ---(6)
Let this be equation (6)
At T = 6 s,
\[\omega =\dfrac{2\pi }{6}\]
\[\omega =\dfrac{\pi }{3}\]---(7)
Let this be equation (7)
Substitute equations (7) and (5) in equation (1)
We get,
\[x=10\sin [(\dfrac{\pi }{3}t)+(\dfrac{\pi }{6})]\]---(8)
Let this be equation (8)
Differentiating equation (8) with respect to time
We get
\[\dfrac{dx}{dt}=10\cos [(\dfrac{\pi }{3}t)+(\dfrac{\pi }{6})](\dfrac{\pi }{3})\]---(9)
 Let this be equation (9)
Again, Differentiating equation (9) with respect to time
We get,
\[\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-10\sin [(\dfrac{\pi }{3}t)+(\dfrac{\pi }{6})]{{(\dfrac{\pi }{3})}^{2}}\]-----(10)
 Let this be equation (10)
\[\dfrac{{{d}^{2}}x}{d{{t}^{2}}}\] is actually the rate of change of velocity.
That is, acceleration
Let acceleration be \[\vec{a}\]
Therefore,
\[\vec{a}=-10\sin [(\dfrac{\pi }{3}t)+(\dfrac{\pi }{6})]{{(\dfrac{\pi }{3})}^{2}}\]---(11)
 Let this be equation (11)
Acceleration at time t = 4 seconds is given by,
\[|\vec{a}|=-10\sin [(\dfrac{\pi }{3}\text{x4})+(\dfrac{\pi }{6})]{{(\dfrac{3.14}{3})}^{2}}\]----(12)
Let this be equation (12)
After calculating the values,
We get,
\[|\vec{a}|=\dfrac{-100}{9}\sin [\dfrac{3\pi }{2}]\] ----(13)
Let this be equation (13)
The value of
\[\sin [\dfrac{3\pi }{2}]\] is -1
Substitute this value in equation (13)
We get,
\[|\vec{a}|=\dfrac{-100}{9}(-1)\]
\[|\vec{a}|=11\]
Therefore, the magnitude of the acceleration of the particle at t = 4 s is
\[|\vec{a}|=11cm/{{s}^{2}}\]

Note:
\[x\text{ }=\text{ }A\text{ }sin\text{ }\left( \omega t\text{ }+\text{ }\Phi \right)\]is the solution for a simple harmonic motion for a particle at any time ‘t’
\[x\text{ }=\text{ }A\text{ }sin\text{ }\left( \omega t \right)\]is the solution for a simple harmonic motion for a particle at its mean position.
\[x\text{ }=\text{ }A\text{ }sin\text{ }\left( \phi \right)\] is the solution for a simple harmonic motion for a particle at the beginning.