1
JEE Main 2022 (Online) 24th June Evening Shift
+4
-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 :

A
1 : 2
B
3 : 2
C
3 : 1
D
2 : 3
2
JEE Main 2022 (Online) 24th June Morning Shift
+4
-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 :

A
2 cm
B
4 cm
C
5.8 cm
D
8 cm
3
JEE Main 2021 (Online) 22th July Evening Shift
+4
-1
The motion of a mass on a spring, with spring constant K is as shown in figure.

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 :
A
$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {2v(0)}}} \right)$$
B
$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {v(0)}}} \right)$$
C
$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$
D
$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$
4
JEE Main 2021 (Online) 17th March Evening Shift
+4
-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?
A
Y(x, t) = 0.03 [ sin 5.1x $$-$$ (0.2 $$\times$$ 103)t ]
B
Y(x, t) = 0.03 [ sin 5.1x $$-$$ (1.5 $$\times$$ 103)t ]
C
Y(x, t) = 0.06 [ sin 5.1x $$-$$ (1.5 $$\times$$ 103)t ]
D
Y(x, t) = 0.06 [ sin 0.8x $$-$$ (0.5 $$\times$$ 103)t ]
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