IIT-JEE 1999 Screening | Simple Harmonic Motion Question 25 | Physics

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