1
MCQ (Single Correct Answer)

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)
2
MCQ (Single Correct Answer)

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
1 rad/s
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.
3
MCQ (Single Correct Answer)

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.
4
MCQ (Single Correct Answer)

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

Questions Asked from Simple Harmonic Motion

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