1

### 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 = $\pm$ ${A \over {\sqrt 2 }}$
2

### 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)
3

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

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $\omega$. If the radius of the bottle is 2.5 cm then $\omega$ is close to – (density of water = 103 kg/m3).
A
2.50 rad s$-$1
B
3.75 rad s$-$1
C
5.00 rad s$-$1
D
7.90 rad s$-$1

## Explanation

Restoring force due to pressing the bottle with small amount x,

F = $- \left( {\rho Ax} \right)g$

$\Rightarrow$ ma = $- \left( {\rho Ax} \right)g$

$\Rightarrow$ a = $- \left( {{{\rho Ag} \over m}} \right)x$

$\therefore$ ${{\omega ^2} = {{\rho Ag} \over m}}$ = ${{{\rho \left( {\pi {r^2}} \right)g} \over m}}$

$\Rightarrow$ $\omega$ = $\sqrt {{{{{10}^3} \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2} \times 10} \over {310 \times {{10}^{ - 3}}}}}$ = 7.90 rad/s
4

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

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -
A
${{4\pi } \over 3}$
B
${3 \over 8}\pi$
C
${7 \over 3}\pi$
D
${{8\pi } \over 3}$

## Explanation

$v = \omega \sqrt {{A^2} - {x^2}} \,\,$    . . .(1)

$a = - {\omega ^2}x$               . . .(2)

$\left| v \right| = \left| a \right|$                   . . .(3)

$\omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x$

${A^2} - {x^2} = {\omega ^2}{x^2}$

${5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)$

$\Rightarrow \,\,\,3 = \omega \times 4$

$T = 2\pi /\omega$