JEE Main 2019 (Online) 12th January Morning Slot | Simple Harmonic Motion Question 68 | Physics

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)

-1

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 ?

The wave is propagating along the positive x-axis with speed 100 ms^–1

The wave is propagating along the positive x-axis with speed 25 ms^–1

The wave is propagating along the negative x-axis with speed 25 ms^–1

The wave is propagating along the negative x-axis with speed 100 ms^–1

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)

-1

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 :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Simple Harmonic Motion Question 68 English

JEE Main 2019 (Online) 12th January Morning Slot Physics - Simple Harmonic Motion Question 68 English

$${1 \over {2\pi }}\sqrt {{{3k} \over m}} $$

$${1 \over {2\pi }}\sqrt {{{6k} \over m}} $$

$${1 \over {2\pi }}\sqrt {{k \over m}} $$

$${1 \over {2\pi }}\sqrt {{{2k} \over m}} $$

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)

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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 :

1 rad/s

10^$$-$$3 rad/s

10^$$-$$1 rad/s

10^$$-$$5 rad/s

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)

-1

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 :

$${K_2}$$ = $${{{K_1}} \over 2}$$

K₂ = 2K₁

K₂ = K₁

K₂ = $${{{K_1}} \over 4}$$

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