1

### 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)

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

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

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

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

$T = 2\pi /\omega$
2

### JEE Main 2019 (Online) 11th January Morning Slot

A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin${{\pi t} \over {90}}$. The ratio of kinetic to potential energy of this particle at t = 210 s will be:
A
${1 \over 9}$
B
3
C
2
D
1

## Explanation

K = ${1 \over 2}$m${\omega ^2}$A2cos2$\omega$t

U = ${1 \over 2}m{\omega ^2}$ A2 sin2 $\omega$t

${k \over U}$ = cot2 $\omega$t = cot2 ${\pi \over {90}}$(210) = ${1 \over 3}$

Hence ratio is 3 (most appropriate)
3

### JEE Main 2019 (Online) 11th January Evening Slot

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :
A
B
10$-$3 rad/s
C
10$-$1 rad/s
D
10$-$5 rad/s

## Explanation

Angular frequency of pendulum

$\omega$ = $\sqrt {{{{g_{eff}}} \over \ell }}$

$\therefore$   ${{\Delta \omega } \over \omega }$ = ${1 \over 2}$ ${{\Delta {g_{eff}}} \over {{g_{eff}}}}$

$\Delta $$\omega = {1 \over 2} {{\Delta g} \over g} \times \omega [{\omega _s} = angular frequency of support] \Delta$$\omega$ = ${1 \over 2} \times {{2A\omega _s^2} \over {100}} \times 100$

$\Delta \omega = {10^{ - 3}}$ rad/sec.
4

### JEE Main 2019 (Online) 11th January Evening Slot

A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then :
A
${K_2}$ = ${{{K_1}} \over 2}$
B
K2 = 2K1
C
K2 = K1
D
K2 = ${{{K_1}} \over 4}$

## Explanation

Maximum kinetic energy at lowest point B is given by

K = mg$\ell$ (1 $-$ cos $\theta$)

where $\theta$ = angular amp.

K1 = mg$\ell$ (1 $-$ cos $\theta$)

K2 = mg(2$\ell$) (1 $-$ cos $\theta$)

K2 = 2K1.