IIT-JEE 2001 Screening | Simple Harmonic Motion Question 23 | Physics

IIT-JEE 2001 Screening

MCQ (Single Correct Answer)

-0.5

A particle executes simple harmonic motion between x = - A to x = + A. The time taken for it to go from 0 to $${A \over 2}$$ is T₁ and to go from $${A \over 2}$$ to A is T₂. Then

T₁ < T₂

T₁ > T₂

T₁ = T₂

T₁ = 2T₂

IIT-JEE 2000 Screening

MCQ (Single Correct Answer)

-0.5

The period of oscillation of a simple pendulum of length $$L$$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $$\alpha$$, is given by

$$2\pi \sqrt {{L \over {g\cos \alpha }}} $$

$$2\pi \sqrt {{L \over {g\sin \alpha }}} $$

$$2\pi \sqrt {{L \over g}} $$

$$2\pi \sqrt {{L \over {g\tan \alpha }}} $$

IIT-JEE 1999 Screening

MCQ (Single Correct Answer)

-0.5

A particle free to move along the x-axis has potential energy given by $$U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]$$ for $$ - \infty \le x \le - \infty $$, where k is a positive constant of appropriate dimensions. Then

at points away from the origin, the particle is in unstable equilibrium

for any finite nonzero value of x, there is a force directed away from the origin

if its total mechanical energy is $${k \over 2}$$, it has its minimum kinetic energy at the origin

for small displacement from x = 0, the motion is simple harmonic

IIT-JEE 1988

MCQ (Single Correct Answer)

-0.25

Two bodies M and N of equal masses are suspended from two separate massless springs of spring constant k₁ and k₂ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is

$${{{k_1}} \over {{k_2}}}$$

$$\sqrt {{{{k_1}} \over {{k_2}}}} $$

$${{{k_2}} \over {{k_1}}}$$

$$\sqrt {{{{k_2}} \over {{k_1}}}} $$

JEE Advanced Subjects