JEE Main 2022 (Online) 26th June Evening Shift | Waves Question 23 | Physics

JEE Main 2022 (Online) 26th June Evening Shift

MCQ (Single Correct Answer)

-1

For a specific wavelength 670 nm of light coming from a galaxy moving with velocity v, the observed wavelength is 670.7 nm. The value of v is :

3 $$\times$$ 10⁸ ms^$$-$$1

3 $$\times$$ 10¹⁰ ms^$$-$$1

3.13 $$\times$$ 10⁵ ms^$$-$$1

4.48 $$\times$$ 10⁵ ms^$$-$$1

JEE Main 2022 (Online) 24th June Evening Shift

MCQ (Single Correct Answer)

-1

Two massless springs with spring constants 2 k and 9 k, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :

1 : 2

3 : 2

3 : 1

2 : 3

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 35 English

JEE Main 2021 (Online) 22th July Evening Shift Physics - Waves Question 35 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 :

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

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

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

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

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