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Question: A particle executes SHM of type \(x=asin\omega t\). It takes time \(t_1\) from x=0 to x=a/2 and \(t_...

A particle executes SHM of type x=asinωtx=asin\omega t. It takes time t1t_1 from x=0 to x=a/2 and t2t_2 from x=a/2 to x=a. The ratio of t1:t2t_1:t_2 will be:
A. 1:1
B. 1:2
C. 1:3
D. 2:1

Explanation

Solution

A motion in which a particle undergoes periodic motion is called Simple harmonic motion (S.H.M). Not every periodic motion is S.H.M but every S.H.M is periodic motion. The revolution of earth about the sun is an example of periodic motion but it is not simple harmonic. A motion is said to be simple harmonic only if the acceleration of the particle is the function of first power of displacement and having direction opposite of the displacement.

Complete answer:
First, let’s understand the standard S.H.M equation.
Y=asin(ωt+ϕ)Y=asin\left( \omega t+\phi \right)is called the standard S.H.M equation.
Here ‘Y’ represents the displacement of wave particles at time ‘t’. Coefficient of trigonometric function ‘a’ is called the amplitude of the wave.’ω\omega’ is the angular frequency of the wave, which is the measure of angular displacement. ‘ϕ\phi’ is the initial phase difference of the wave. It is also called ‘epoch’.
In the equation given in the question: x=asinωtx=asin\omega t, ϕ\phi=0.
Now, given the times taken by the body to move from x=0 to x=a/2 is t2t_2.
Thus, putting the values in the equation: x=asinωtx=asin\omega t, we get;
a2=asinωt1\dfrac a2 = asin\omega t_1
    sinωt1=12\implies sin\omega t_1 = \dfrac12
    ωt1=sin112=π6\implies \omega t_1 = sin^{-1} \dfrac12 = \dfrac{\pi}{6}
Thus t1=π6ωt_1 = \dfrac{\pi}{6\omega}
Also, the time taken to reach from x=a/2 to x=a is t2t_2.
As we know that the time to reach from x=0 to x=a is T4=π2ω\dfrac T4 = \dfrac{\pi}{2\omega} [as ω=2πω\omega = \dfrac{2\pi}\omega]
Hence, t2=π2ωπ6ω=π3ωt_2 = \dfrac{\pi}{2\omega}-\dfrac{\pi}{6\omega}=\dfrac{\pi}{3\omega}
Thus t1t2=π6ωπ3ω=12\dfrac{t_1}{t_2} = \dfrac{\dfrac{\pi}{6\omega}}{\dfrac{\pi}{3\omega}}=\dfrac12

So, the correct answer is “Option B”.

Note:
Students must not get confused by the actual phase and initial phase ( phase at t=0 ). Actual phase means phase difference between the two particles at any general time i.e. the whole term inside of trigonometric functions. Eg sin(ωt+ϕ\omega t+\phi), ωt+ϕ\omega t+\phi is the actual phase of the particle.