The displacement of a particle executing simple harmonic motion is given by
$y = A_{0} + A sin ω t + B cos ω t$ . Then, the amplitude of its oscillation is given by

Question

The displacement of a particle executing simple harmonic motion is given by y - A-0 - A sin omega t - B cos omega tThen the amplitude of its oscillation is given by -

Toppr Admin · Accepted Answer

Correct option is B- -sqrt-A-2 - B-2-Given-y - A-0 - A sin -omega t - B sin -omega t-Equation of-160- SHM-y- - y - A-0 - A sin -omega-160- t - B cos -omega t-Resultant amplitude-R - -sqrt-A-2 - B-2 - 2 AB cos 90-circ- -sqrt-A-2 - B-2-

The displacement of a particle executing simple harmonic motion is given by $$y = A_0 + A \sin \omega t + B \cos \omega t$$
Then the amplitude of its oscillation is given by :

Correct option is B. $$\sqrt{A^2 + B^2}$$
Given,
$$y = A_0 + A sin \omega t + B sin \omega t$$
Equation of SHM
$$y' = y - A_0 = A sin \omega t + B cos \omega t$$
Resultant amplitude,
$$R = \sqrt{A^2 + B^2 + 2 AB cos 90^{\circ}}$$
$$= \sqrt{A^2 + B^2}$$

The displacement of a particle executing simple harmonic motion is given by $$y = A_0 + A \sin \omega t + B \cos \omega t$$Then the amplitude of its oscillation is given by :

Correct option is B. $$\sqrt{A^2 + B^2}$$Given,$$y = A_0 + A sin \omega t + B sin \omega t$$Equation of SHM$$y' = y - A_0 = A sin \omega t + B cos \omega t$$Resultant amplitude,$$R = \sqrt{A^2 + B^2 + 2 AB cos 90^{\circ}}$$$$= \sqrt{A^2 + B^2}$$

The displacement of a particle executing simple harmonic motion is given by $$y = A_0 + A \sin \omega t + B \cos \omega t$$
Then the amplitude of its oscillation is given by :

Correct option is B. $$\sqrt{A^2 + B^2}$$
Given,
$$y = A_0 + A sin \omega t + B sin \omega t$$
Equation of SHM
$$y' = y - A_0 = A sin \omega t + B cos \omega t$$
Resultant amplitude,
$$R = \sqrt{A^2 + B^2 + 2 AB cos 90^{\circ}}$$
$$= \sqrt{A^2 + B^2}$$