1

### JEE Main 2018 (Online) 16th April Morning Slot

An oscillator of mass M is at rest in its equilibrium position in a potential
V = ${1 \over 2}$ k(x $-$ X)2. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1)
A
${1 \over {\sqrt 3 }}$
B
${1 \over 2}$
C
${2 \over 3}$
D
${3 \over {\sqrt 5 }}$
2

### JEE Main 2018 (Online) 16th April Morning Slot

A particle executes simple harmonic motion and is located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The freqquency of the oscillation is :
A
${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)$
B
${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)$
C
${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)$
D
${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)$

## Explanation

In general equation of simple harmonic motion, y = A sin $\omega$t

$\therefore\,\,\,$ a = A sin $\omega$t0

$\,\,\,\,\,\,$ b = A sin 2$\omega$t0

$\,\,\,\,\,\,\,$c = A sin 3$\omega$t0

a + c = A[sin $\omega$t0 + sin 3$\omega$t0]

= 2A sin 2$\omega$t0 cos$\omega$t0

$\Rightarrow $$\,\,\,a + c = 2 b cos\omega t0 \Rightarrow$$\,\,\,$ ${{a + c} \over b}$ = 2 cos$\omega$t0

$\Rightarrow$$\,\,\,$ $\omega$ = ${1 \over {{t_0}}}$ cos$-$1 $\left( {{{a + c} \over {2b}}} \right)$

$\therefore\,\,\,$ f = ${\omega \over {2\pi }}$

= ${1 \over {2\pi {t_0}}}$ cos$-$1 $\left( {{{a + c} \over {2b}}} \right)$
3

### JEE Main 2019 (Online) 9th January Evening Slot

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :
A
${A \over 2}$
B
${A \over {2\sqrt 2 }}$
C
${A \over {\sqrt 2 }}$
D
A

## Explanation

Total energy of particle = ${1 \over 2}k{A^2}$

Potential energy (v) = ${1 \over 2}$ kx2

Kinetic energy (K) = ${1 \over 2}$ kA2 $-$ ${1 \over 2}$kx2

According to the question,

Potential energy = Kinetic energy

$\therefore$  ${1 \over 2}$kx2 = ${1 \over 2}$kA2 $-$ ${1 \over 2}$ kx2

$\Rightarrow$  kx2 = ${1 \over 2}$ kA2

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

### JEE Main 2019 (Online) 10th January Evening Slot

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz)
A
4
B
7
C
6
D
5

## Explanation

For closed organ pipe, resonate frequency is odd multiple of fundamental frequency.

$\therefore$  (2n + 1) f0 $\le$ 20,000

(f0 is fundamental frequency = 1.5 KHz)

$\therefore$  n = 6

$\therefore$  Total number of overtone that can be heared is 7. (0 to 6)