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

$A$ Carnot engine operating between temperatures $T_1$ and $T_2$ has an efficiency of $\frac{1}{6}$. When $T_2$ is lowered by $62\, K$,its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are,respectively:

Which of the following is not a sex hormone?

If $\int \sec ^2 x \operatorname{cosec}^4 x \, dx = -\frac{1}{3} \cot ^3 x + k \tan x - 2 \cot x + C$, then $k$ is equal to

The major product obtained in the following reaction is

Which is the smallest angiosperm?

Vedclass Test Series

Exam Paper Generator

Online Exam Module