1

### 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.
2

### JEE Main 2019 (Online) 12th January Morning Slot

A travelling harmonic wave is represented by the equation y(x,t) = 10–3sin (50t + 2x), where, x and y are in mater and t is in seconds. Which of the following is a correct statement about the wave ?
A
The wave is propagating along the positive x-axis with speed 100 ms–1
B
The wave is propagating along the positive x-axis with speed 25 ms–1
C
The wave is propagating along the negative x-axis with speed 25 ms–1
D
The wave is propagating along the negative x-axis with speed 100 ms–1

## Explanation

y = a sin($\omega$t + kx)

$\Rightarrow$  wave is moving along $-$ve x-axis with speed

v = ${\omega \over K}$ $\Rightarrow$  v = ${{50} \over 2}$ = 25m/sec
3

### JEE Main 2019 (Online) 12th January Morning Slot

Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length $\ell$ and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is : A
${1 \over {2\pi }}\sqrt {{{3k} \over m}}$
B
${1 \over {2\pi }}\sqrt {{{6k} \over m}}$
C
${1 \over {2\pi }}\sqrt {{k \over m}}$
D
${1 \over {2\pi }}\sqrt {{{2k} \over m}}$

## Explanation $\tau = - 2kx{\ell \over 2}\cos \theta$

$\Rightarrow$   $\tau = \left( {{{K{\ell ^2}} \over 2}} \right)\theta = - C\theta$

$\Rightarrow$   $f = {1 \over {2\pi }}\sqrt {{C \over 1}} = {1 \over {2\pi }}\sqrt {{{{{K{\ell ^2}} \over 2}} \over {{{M{\ell ^2}} \over {12}}}}}$

$\Rightarrow$    $f = {1 \over {2\pi }}\sqrt {{{6K} \over M}}$
4

### JEE Main 2019 (Online) 12th January Evening Slot

A simple harmonic motion is represented by :

y = 5 (sin 3 $\pi$ t + $\sqrt 3$ cos 3 $\pi$t) cm

The amplitude and time period of the motion are :
A
10 cm, ${3 \over 2}$ s
B
5 cm, ${2 \over 3}$ s
C
5 cm, ${3 \over 2}$ s
D
10 cm, ${2 \over 3}$ s

## Explanation y = 5[sin(3$\pi$t) + $\sqrt 3$cos(3$\pi$t)]

= 10sin $\left( {3\pi t + {\pi \over 3}} \right)$

Amplitude = 10 cm

T = ${{2\pi } \over w}$ = ${{2\pi } \over {3\pi }}$ = ${2 \over 3}$ sec