1
IIT-JEE 2011 Paper 2 Offline
+2
-0.5
A wooden block performs $$SHM$$ on a frictionless surface with frequency, $${v_0}.$$ The block carries a charge $$+Q$$ on its surface . If now a uniform electric field $$\overrightarrow E$$ is switched- on as shown, then the $$SHM$$ of the block will be A
of the same frequency and with shifted mean position.
B
of the same frequency and with the same mean position
C
of changed frequency and with shifted mean position.
D
of changed frequency and with the same mean position.
2
IIT-JEE 2010 Paper 1 Offline
+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). 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
3
IIT-JEE 2010 Paper 1 Offline
+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). 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
$${1 \over A}\sqrt {\alpha /m}$$
4
IIT-JEE 2010 Paper 1 Offline
+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). 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.
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