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### JEE Main 2016 (Online) 10th April Morning Slot

A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to :
A
680 Hz
B
510 Hz
C
340 Hz
D
170 Hz

## Explanation Let the frequency of the horn = f

Apparent frequency heared by the observer directly,

${f_{dir}} = \left( {{V \over {V - {V_s}}}} \right)f = \left( {{{340} \over {340 - 5}}} \right)f = {{340} \over {335}}f$

Apparent frequency heared by the observer on reflection from the wall,

${f_{ind}} = \left( {{V \over {V + {V_s}}}} \right)f = \left( {{{340} \over {340 + 5}}} \right)f = {{340} \over {345}}f$

Also, given that,

find $-$ fdir $=$ 5

$\Rightarrow$   ${{340f} \over {345}}$ $-$ ${{340f} \over {335}}$ $=$ 5

$\Rightarrow$   f $=$ ${5 \over {340}} \times {{335 \times 345} \over {10}}$

= 169.9  $\simeq$   170 Hz
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### JEE Main 2017 (Offline)

A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy – time graph of the particle will look like:
A B C D ## Explanation

For a particle executing SHM

At mean position; t = 0, $\omega$t = 0, y = 0, V = Vmax = a$\omega$

$\therefore$ K.E = K.Emax = ${1 \over 2}m{\omega ^2}{a^2}$

At extreme position : t = ${T \over 4}$ , $\omega$t = ${\pi \over 2}$ , y = A, V = Vmin = 0

$\therefore$ K.E = K.Emin = 0

Hence graph (A) correctly represents kinetic energy - time graph.
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### JEE Main 2017 (Online) 8th April Morning Slot

Two wires W1 and W2 have the same radius r and respective densities $\rho$1 and $\rho$2 such that ρ2 = 4$\rho$1 . They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W1 to W2 is : A
1 : 1
B
1 : 2
C
1 : 3
D
4 : 1

## Explanation

Fundamental frequency of a stretched string is ,

f = ${n \over {2L}}\sqrt {{T \over \mu }}$

Here, n = number of antinodes

$\mu$ = mass per unit length.

As, O is the midpoint of two bridegs, hence length of two wires are equal.

$\therefore\,\,\,$ L1 = L2 = L

As frequency of both wires same

f1 = f2

$\Rightarrow $$\,\,\, {{{n_1}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _1}}}} = {{{n_2}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _2}}}} \Rightarrow$$\,\,\,$ ${{{n_1}} \over {{n_2}}}$ = $\sqrt {{{{\rho _1}} \over {{\rho _2}}}}$

$\Rightarrow$$\,\,\,$ ${{{n_1}} \over {{n_2}}}$ = $\sqrt {{{{\rho _1}} \over {4{\rho _1}}}}$ = ${1 \over 2}$
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### JEE Main 2017 (Online) 9th April Morning Slot

A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by

y(x, t) = 0.5 sin $\left( {{{5\pi } \over 4}x} \right)\,$ cos(200 $\pi$t).

What is the speed of the travelling wave moving in the positive x direction ?

(x and t are in meter and second, respectively.)
A
160 m/s
B
90 m/s
C
180 m/s
D
120 m/s

## Explanation

Standard equation of standing wave,

y(x, t) = 2a sin kx cos $\omega$t

Given,

y(x, t) = 0.5 sin $\left( {{{5\pi } \over 4}x} \right)$ cos (200$\pi$t).

So, k = ${{5\pi } \over 4}$ and $\omega$ = 200$\pi$

$\therefore\,\,\,$ Speed of travelling wave

= ${\omega \over k}$ = ${{200\pi } \over {{{5\pi } \over 4}}}$ = 160 m/s.

NEET