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1

### JEE Main 2021 (Online) 22th July Evening Shift

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

## Explanation

$$C\cos \phi = x(0)$$

$$tC\omega \sin \phi = v(0)$$

$${\left[ {{{v(0)} \over \omega }} \right]^2} + {[x(0)]^2} = {C^2}$$

$$\tan \phi = {{v(0)} \over {x(0)\omega }}$$
2

### JEE Main 2021 (Online) 17th March Evening Shift

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 ]

## Explanation

$$Y = A\sin (kx - \omega t)$$

$$A = {6 \over 2}$$ = 3cm = 0.03 m

$$\omega = 2\pi f = 2\pi \times 245$$

$$\omega = 1.5 \times {10^3}$$

$$k = {\omega \over v} = {{1.5 \times {{10}^3}} \over {300}}$$

$$k = 5.1$$

$$y = 0.03\sin (5.1x - (1.5 \times {10^3})t)$$
3

### JEE Main 2021 (Online) 17th March Morning Shift

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ?
A
x = $${A \over 2}$$
B
x = $$\pm$$ A
C
x = $$\pm$$ $${A \over {\sqrt 2 }}$$
D
x = 0

## Explanation

KE = PE

$${1 \over 2}k({A^2} - {X^2}) = {1 \over 2}K{X^2}$$

$$\Rightarrow$$ $${A^2} - {X^2} = {X^2}$$

$$\Rightarrow$$ $$2{X^2} = {A^2}$$

$$\Rightarrow$$ $${X^2} = {{{A^2}} \over {\sqrt 2 }}$$

$$\Rightarrow$$ $$X = \pm {A \over {\sqrt 2 }}$$
4

### JEE Main 2021 (Online) 26th February Evening Shift

A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is field, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?
A
335 Hz
B
345 Hz
C
338 Hz
D
342 Hz

## Explanation

Initially beat frequency = 5Hz

so, $$\rho$$A = 340 $$\pm$$ 5 = 345 Hz, or 335 Hz

after filing frequency increases slightly so, new value of frequency of A > $$\rho$$A

Now, beat frequency = 2Hz

$$\Rightarrow$$ new $$\rho$$A = 340 $$\pm$$ 2 = 342 Hz, or 338 Hz

hence, original frequency of A is $$\rho$$A = 335 Hz

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