Two simple harmonic motions are represented b | vedclass

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

Question

(B) The first equation is $y_{1} = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$.We can rewrite this by multiplying and dividing by $2$: $y_{1} = 10 [\frac

Vedclass · Accepted Answer

$2: 1$

Vedclass · Answer

(B) The first equation is $y_{1} = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$.We can rewrite this by multiplying and dividing by $2$: $y_{1} = 10 [\frac{1}{2} \sin 2 \pi t + \frac{\sqrt{3}}{2} \cos 2 \pi t]$.Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we set $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.Thus,$y_{1} = 10 [\sin 2 \pi t \cos \frac{\pi}{3} + \cos 2 \pi t \sin \frac{\pi}{3}] = 10 \sin(2 \pi t + \frac{\pi}{3})$.Therefore,the amplitude $A_{1} = 10$.For the second equation,$y_{2} = 5 \sin (2 \pi t + \frac{\pi}{4})$,the amplitude $A_{2} = 5$.The ratio of the amplitudes is $\frac{A_{1}}{A_{2}} = \frac{10}{5} = \frac{2}{1}$.

Two simple harmonic motions are represented by $y_{1} = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$ and $y_{2} = 5 \sin (2 \pi t + \frac{\pi}{4})$. The ratio of their amplitudes is

Similar Questions

The three-domain system of classification is based on which of the following criteria?

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

Rutherford’s scattering experiment is related to the size of the

$A$ ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation,which of the following is true?

The epithelial tissue present on the inner surface of bronchioles and fallopian tubes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module