The motion of a particle represented by y = s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equation: $y = \sin \omega t - \cos \omega t$Multiply and divide by $\sqrt{2}$:$y = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega t - \frac

Vedclass · Accepted Answer

$S.H.M.$ with period $\frac{2\pi}{\omega}$,amplitude $\sqrt{2}$

Vedclass · Answer

(B) Given equation: $y = \sin \omega t - \cos \omega t$Multiply and divide by $\sqrt{2}$:$y = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t \right)$Using trigonometric identities $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:$y = \sqrt{2} \left( \sin \omega t \cos \frac{\pi}{4} - \cos \omega t \sin \frac{\pi}{4} \right)$Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$y = \sqrt{2} \sin \left( \omega t - \frac{\pi}{4} \right)$This is the standard equation of $S.H.M.$ of the form $y = A \sin(\omega t + \phi)$,where amplitude $A = \sqrt{2}$ and angular frequency is $\omega$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.

The motion of a particle represented by $y = \sin \omega t - \cos \omega t$ is

Similar Questions

$A$ body is projected up with a velocity equal to $\frac{3}{4}$ of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth $= R$)

When a capacitor is connected in series to an $LR$ circuit,the alternating current flowing in the circuit

The value of the series $\cos 12^{\circ} + \cos 84^{\circ} + \cos 132^{\circ} + \cos 156^{\circ}$ is

Which of the following is $NOT$ an effect of the filarial worm?

The diameter of a plano-convex lens is $6 \ cm$ and its thickness at the center is $3 \ mm$. If the speed of light in the material of the lens is $2 \times 10^8 \ m/s$,the focal length of the lens is.......$cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module