
A particle executes SHM of type $x=asin\omega t$. It takes time $t_1$ from x=0 to x=a/2 and $t_2$ from x=a/2 to x=a. The ratio of $t_1:t_2$ will be:
A. 1:1
B. 1:2
C. 1:3
D. 2:1
Answer
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Hint: 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.
$Y=asin\left( \omega t+\phi \right)$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.’$\omega$’ is the angular frequency of the wave, which is the measure of angular displacement. ‘$\phi$’ is the initial phase difference of the wave. It is also called ‘epoch’.
In the equation given in the question: $x=asin\omega t$, $\phi$=0.
Now, given the times taken by the body to move from x=0 to x=a/2 is $t_2$.
Thus, putting the values in the equation: $x=asin\omega t$, we get;
$\dfrac a2 = asin\omega t_1$
$\implies sin\omega t_1 = \dfrac12$
$\implies \omega t_1 = sin^{-1} \dfrac12 = \dfrac{\pi}{6}$
Thus $t_1 = \dfrac{\pi}{6\omega}$
Also, the time taken to reach from x=a/2 to x=a is $t_2$.
As we know that the time to reach from x=0 to x=a is $\dfrac T4 = \dfrac{\pi}{2\omega}$ [as $\omega = \dfrac{2\pi}\omega$]
Hence, $t_2 = \dfrac{\pi}{2\omega}-\dfrac{\pi}{6\omega}=\dfrac{\pi}{3\omega}$
Thus $\dfrac{t_1}{t_2} = \dfrac{\dfrac{\pi}{6\omega}}{\dfrac{\pi}{3\omega}}=\dfrac12$
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($\omega t+\phi$), $\omega t+\phi$ is the actual phase of the particle.
Complete answer:
First, let’s understand the standard S.H.M equation.
$Y=asin\left( \omega t+\phi \right)$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.’$\omega$’ is the angular frequency of the wave, which is the measure of angular displacement. ‘$\phi$’ is the initial phase difference of the wave. It is also called ‘epoch’.
In the equation given in the question: $x=asin\omega t$, $\phi$=0.
Now, given the times taken by the body to move from x=0 to x=a/2 is $t_2$.
Thus, putting the values in the equation: $x=asin\omega t$, we get;
$\dfrac a2 = asin\omega t_1$
$\implies sin\omega t_1 = \dfrac12$
$\implies \omega t_1 = sin^{-1} \dfrac12 = \dfrac{\pi}{6}$
Thus $t_1 = \dfrac{\pi}{6\omega}$
Also, the time taken to reach from x=a/2 to x=a is $t_2$.
As we know that the time to reach from x=0 to x=a is $\dfrac T4 = \dfrac{\pi}{2\omega}$ [as $\omega = \dfrac{2\pi}\omega$]
Hence, $t_2 = \dfrac{\pi}{2\omega}-\dfrac{\pi}{6\omega}=\dfrac{\pi}{3\omega}$
Thus $\dfrac{t_1}{t_2} = \dfrac{\dfrac{\pi}{6\omega}}{\dfrac{\pi}{3\omega}}=\dfrac12$
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($\omega t+\phi$), $\omega t+\phi$ is the actual phase of the particle.
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