1

### 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.
2

### JEE Main 2018 (Offline)

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 1023 gm mole–1)
A
5.5 N/m
B
6.4 N/m
C
7.1 N/m
D
2.2 N/m

## Explanation

6.02 $\times$ 1023 atoms of silver = 108 gm

1   atoms   of   silver   =   ${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$ kg

For a harmonic oscillator

f = ${1 \over {2\pi }}$ $\sqrt {{k \over m}}$

Where  k = force constant

$\Rightarrow $$\,\,\, f2 = {1 \over {4{\pi ^2}}} \left( {{k \over m}} \right) \Rightarrow$$\,\,\,$ k = mf2 $\times$ 4$\pi$2

Given,

f = 1012

m = ${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$

$\therefore\,\,\,$ k = ${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$ $\times$ 1012 $\times$ 4$\pi$2

= 7.1 N/m
3

### JEE Main 2018 (Offline)

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 $\times$ 103 kg/m3 and its Young’s modulus is 9.27 $\times$ 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?
A
7.5 kHz
B
5 kHz
C
2.5 kHz
D
10 kHz

## Explanation

As   rod length = 60 cm

$\therefore\,\,\,$ ${\lambda \over 2}$ = 60

$\Rightarrow $$\,\,\, \lambda = 120 cm = 1.2 m In solid, velocity of wave, V = \sqrt {{Y \over \rho }} = \sqrt {{{9.27 \times {{10}^{10}}} \over {2.7 \times {{10}^3}}}} = 5.85 \times 103 m/sec. As we know, v = f \lambda \therefore\,\,\, f = {v \over \lambda } = {{5.85 \times {{10}^3}} \over {1.2}} = 4.88 \times 103 Hz \simeq 5 kHz 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Morning Slot A tuning fork vibrates with frequency 256 Hz and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe ? (Speed of sound in air is 340\,m{s^{ - 1}}) A 220 cm B 190 cm C 180 cm D 200 cm ## Explanation The tuning fork vibrates with frequency 256 Hz and give one beat per second So, the organ pipe will have frequency (256 \pm 1) Hr. For open organ pipe, Frequency n = {{N\upsilon } \over {2\ell }} Here n = 255 Hz N = 3 \upsilon = 340 m/s \therefore\,\,\,\, 255 = {{3 \times 340} \over {2 \times \ell }} \Rightarrow$$\,\,\,\,$ $\ell$ = ${{3 \times 340} \over {2 \times 255}} = 2\,m$

$\therefore\,\,\,\,$ $\ell$ = 2m or 200 cm