(D) The phase of the first particle is $\theta_1 = \frac{\pi}{2} t + \phi$.At $t = 1 \, s$,$\theta_1 = \frac{\pi}{2}(1) + \phi = \frac{\pi}{2} + \phi$

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(D) The phase of the first particle is $\theta_1 = \frac{\pi}{2} t + \phi$.At $t = 1 \, s$,$\theta_1 = \frac{\pi}{2}(1) + \phi = \frac{\pi}{2} + \phi$.The phase of the second particle is $\theta_2 = \frac{2 \pi}{3} t + \phi$.At $t = 1 \, s$,$\theta_2 = \frac{2 \pi}{3}(1) + \phi = \frac{2 \pi}{3} + \phi$.The phase difference $\Delta \theta$ is given by $|\theta_2 - \theta_1|$.$\Delta \theta = |(\frac{2 \pi}{3} + \phi) - (\frac{\pi}{2} + \phi)| = |\frac{2 \pi}{3} - \frac{\pi}{2}| = |\frac{4 \pi - 3 \pi}{6}| = \frac{\pi}{6}$.

Class 11 Physics Oscillations Position of a Particle in SHM, Displacement and Phase

Easy

The displacements of two particles executing $S.H.M.$ on the same line are given as $y_1 = a \sin \left(\frac{\pi}{2} t + \phi\right)$ and $y_2 = b \sin \left(\frac{2 \pi}{3} t + \phi\right)$. The phase difference between them at $t = 1 \, s$ is .........

A
$\pi$
B
$\frac{\pi}{2}$
C
$\frac{\pi}{4}$
D
$\frac{\pi}{6}$

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Oscillations

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Position of a Particle in SHM, Displacement and Phase

$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

MediumAIPMT 2007

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After the beginning of motion,how long will it take for a harmonically oscillating particle to reach a displacement equal to one-half of its amplitude,if the time period is $24 \ sec$ and the particle starts from rest?

Easy

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The phase difference between two $SHM$ equations $y_1 = 10 \sin(10\pi t + \frac{\pi}{3})$ and $y_2 = 12 \sin(8\pi t + \frac{\pi}{4})$ at $t = 0.5 \ s$ is:

Medium

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$A$ particle executing a simple harmonic motion has a period of $6 \,s$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

DifficultKCET 2011

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$A$ simple harmonic oscillator has an amplitude $a$ and time period $T$. The time required by it to travel from $x = a$ to $x = a/2$ is

MediumAIPMT 1992

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The displacements of two particles executing $S.H.M.$ on the same line are given as $y_1 = a \sin \left(\frac{\pi}{2} t + \phi\right)$ and $y_2 = b \sin \left(\frac{2 \pi}{3} t + \phi\right)$. The phase difference between them at $t = 1 \, s$ is .........

Similar Questions

$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

After the beginning of motion,how long will it take for a harmonically oscillating particle to reach a displacement equal to one-half of its amplitude,if the time period is $24 \ sec$ and the particle starts from rest?

The phase difference between two $SHM$ equations $y_1 = 10 \sin(10\pi t + \frac{\pi}{3})$ and $y_2 = 12 \sin(8\pi t + \frac{\pi}{4})$ at $t = 0.5 \ s$ is:

$A$ particle executing a simple harmonic motion has a period of $6 \,s$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

$A$ simple harmonic oscillator has an amplitude $a$ and time period $T$. The time required by it to travel from $x = a$ to $x = a/2$ is

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