(D) The general equation for $S.H.M.$ is given by $y = A \sin(\omega t + \phi)$.Given amplitude $A = 0.5\, cm$ and time period $T = 0.4\, s$.The angul

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(D) The general equation for $S.H.M.$ is given by $y = A \sin(\omega t + \phi)$.Given amplitude $A = 0.5\, cm$ and time period $T = 0.4\, s$.The angular frequency $\omega$ is calculated as $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.4} = 5\pi\, rad/s$.The initial phase $\phi = \pi/2$.Substituting these values into the general equation:$y = 0.5 \sin(5\pi t + \pi/2)$.Using the trigonometric identity $\sin(\theta + \pi/2) = \cos(\theta)$,we get:$y = 0.5 \cos(5\pi t)$.

Class 11 Physics Oscillations Periodic, Oscillatory motion and its characteristics and types of SHM and Equation of SHM

Medium

The amplitude and the time period in a $S.H.M.$ are $0.5\, cm$ and $0.4\, s$ respectively. If the initial phase is $\pi/2$ radian,then the equation of $S.H.M.$ will be:

A
$y = 0.5\, \sin(5\pi t)$
B
$y = 0.5\, \sin(4\pi t)$
C
$y = 0.5\, \sin(2.5\pi t)$
D
$y = 0.5\, \cos(5\pi t)$

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Periodic, Oscillatory motion and its characteristics and types of SHM and Equation of SHM

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The amplitude and the time period in a $S.H.M.$ are $0.5\, cm$ and $0.4\, s$ respectively. If the initial phase is $\pi/2$ radian,then the equation of $S.H.M.$ will be:

Similar Questions

$A$ ball is released from height $h$ which makes a perfectly elastic collision with the ground. The frequency of the periodic vibratory motion is ($g$ = acceleration due to gravity).

Why is the periodic time $T = \frac{2\pi}{\omega}$ taken for the function $f(t) = A \cos \omega t$?

$A$ potential is given by $V(x) = k(x+a)^2 / 2$ for $x < 0$ and $V(x) = k(x-a)^2 / 2$ for $x > 0$. The schematic variation of the oscillation period $T$ for a particle performing periodic motion in this potential as a function of its energy $E$ is:

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