JEE Main 2022 (Online) 27th June Morning Shift | Waves Question 40 | Physics

JEE Main 2022 (Online) 27th June Morning Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be :

20%

10%

JEE Main 2022 (Online) 24th June Morning Shift

MCQ (Single Correct Answer)

-1

The equations of two waves are given by :

y₁ = 5 sin 2$$\pi$$(x - vt) cm

y₂ = 3 sin 2$$\pi$$(x $$-$$ vt + 1.5) cm

These waves are simultaneously passing through a string. The amplitude of the resulting wave is :

2 cm

4 cm

5.8 cm

8 cm

JEE Main 2021 (Online) 22th July Evening Shift

MCQ (Single Correct Answer)

-1

The motion of a mass on a spring, with spring constant K is as shown in figure.

JEE Main 2021 (Online) 22th July Evening Shift Physics - Waves Question 51 English

JEE Main 2021 (Online) 22th July Evening Shift Physics - Waves Question 51 English

The equation of motion is given by
x(t) = A sin$$\omega$$t + B cos$$\omega$$t with $$\omega$$ = $$\sqrt {{K \over m}} $$

Suppose that at time t = 0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t) = C cos($$\omega$$t $$-$$ $$\phi$$), where C and $$\phi$$ are :

$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {2v(0)}}} \right)$$

$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {v(0)}}} \right)$$

$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$

$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$

JEE Main 2021 (Online) 17th March Evening Shift

MCQ (Single Correct Answer)

-1

A sound wave of frequency 245 Hz travels with the speed of 300 ms^$$-$$1 along the positive x-axis. Each point of the wave moves to and from through a total distance of 6 cm. What will be the mathematical expression of this travelling wave?

Y(x, t) = 0.03 [ sin 5.1x $$-$$ (0.2 $$\times$$ 10³)t ]

Y(x, t) = 0.03 [ sin 5.1x $$-$$ (1.5 $$\times$$ 10³)t ]

Y(x, t) = 0.06 [ sin 5.1x $$-$$ (1.5 $$\times$$ 10³)t ]

Y(x, t) = 0.06 [ sin 0.8x $$-$$ (0.5 $$\times$$ 10³)t ]

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