1
IIT-JEE 2010 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

IIT-JEE 2010 Paper 1 Offline Physics - Simple Harmonic Motion Question 12 English Comprehension

If the total energy of the particle is E, it will perform periodic motion only if

A
E < 0
B
E > 0
C
V0 > E > 0
D
E > V0
2
IIT-JEE 2010 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

IIT-JEE 2010 Paper 1 Offline Physics - Simple Harmonic Motion Question 11 English Comprehension

For periodic motion of small amplitude A, the time period T of this particle is proportional to

A
$$A\sqrt {m/\alpha } $$
B
$${1 \over A}\sqrt {m/\alpha } $$
C
$$A\sqrt {\alpha /m} $$
D
$${1 \over A}\sqrt {\alpha /m} $$
3
IIT-JEE 2010 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

IIT-JEE 2010 Paper 1 Offline Physics - Simple Harmonic Motion Question 10 English Comprehension

The acceleration of this particle for $$|x| > {X_0}$$ is

A
proportional to V0.
B
proportional to V0/mX0.
C
proportional to $$\sqrt {{V_0}/m{X_0}} $$.
D
zero.
4
IIT-JEE 2009 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

The mass M shown in the figure below oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is

IIT-JEE 2009 Paper 2 Offline Physics - Simple Harmonic Motion Question 5 English

A
$${{{k_1}A} \over {{k_2}}}$$
B
$${{{k_2}A} \over {{k_1}}}$$
C
$${{{k_1}A} \over {{k_1} + {k_2}}}$$
D
$${{{k_2}A} \over {{k_1} + {k_2}}}$$
JEE Advanced Subjects
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12