JEE Advanced PYQ Damped, Forced Oscillations and Resonance

Q 1 :

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5 s. In another 10 s it will decrease to $α$ times its original magnitude, where $α$ equals [2013]

0.7
0.81
0.729
0.6

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(3)

$∵$ $A = A_{0} e^{- \frac{b t}{2 m}}$

$(where, A_{0} = maximum amplitude)$

According to the question, after $5$ second,

$0.9 A_{0} = A_{0} e^{- \frac{b (5)}{2 m}} ... (i)$

After $10$ more second,

$α A_{0} = A_{0} e^{- \frac{b (15)}{2 m}} ... (ii)$

From $e q^{n} s$ (i) and (ii)

$α = 0.729$

Q 2 :

A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector $\vec{a}$ is correctly shown in: [2002]

[IMAGE 550]
[IMAGE 551]
[IMAGE 552]
[IMAGE 553]

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(3)

[IMAGE 554]

The resultant of transverse ${\vec{a}}_{t}$ and radial ${\vec{a}}_{r}$ component of the acceleration is represented by $\vec{a}$

Q 3 :

A particle executes simple harmonic motion between $x = - A$ and $x = + A .$ The time taken for it to go from 0 to $\frac{A}{2}$ is $T_{1}$ and to go from $\frac{A}{2}$ to $A$ is $T_{2}$ . Then [2001]

$T_{1} < T_{2}$
$T_{1} > T_{2}$
$T_{1} = T_{2}$
$T_{1} = 2 T_{2}$

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(1)

In SHM, velocity of particles goes on decreasing from maximum value to zero as the particles travel from mean position to extreme position.

Therefore if the time taken for the body to go from $O$ to $\frac{A}{2}$ is $T_{1}$ and to go from $\frac{A}{2}$ to $A$ is $T_{2}$ then obviously $T_{1} < T_{2}$

Q 4 :

A $0.1 kg$ mass is suspended from a wire of negligible mass. The length of the wire is $1 m$ and its cross-sectional area is $4.9 \times 10^{- 7} m^{2} .$ If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140 {rad s}^{- 1} .$ If the Young's modulus of the material of the wire is $n \times 10^{9} {Nm}^{- 2},$ the value of $n$ is ___. [2010]

(4)

$As we know,$

$Y = \frac{F L}{A l} \Rightarrow F = (\frac{Y A}{L}) l$

Also, $F = K l$

or, $K l = (\frac{Y A}{L}) l \Rightarrow K = \frac{Y A}{L}$

Angular frequency $ω = \sqrt{\frac{K}{m}}$ or, $ω = \sqrt{\frac{Y A}{m L}}$

or, $140 = \sqrt{\frac{n \times 10^{9} \times 4.9 \times 10^{- 7}}{0.1 \times 1}}$ $(∵ Y = 9 \times 10^{9} given)$

$∴$ $n = 4$

Q 5 :

A metal rod of length ‘L’ and mass ‘m’ is pivoted at one end. A thin disc of mass ‘M’ and radius ‘R’ (< L) is attached at its center to the free end of the rod. Consider two ways the disc is attached:

(case A) The disc is not free to rotate about its centre and

(case B) the disc is free to rotate about its centre.

The rod-disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is (are) true? [2011]

[IMAGE 555]

Restoring torque in case A = Restoring torque in case B
Restoring torque in case A < Restoring torque in case B
Angular frequency for case A > angular frequency for case B.
Angular frequency for case A < angular frequency for case B.

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Select one or more options

(1, 4)

Applying, torque $τ = I α$

For case A: $m g (\frac{ℓ}{2} \sin θ) + M g (ℓ \sin θ) = I_{A} α_{A} ...(i)$

[IMAGE 556]

For case B: $m g (\frac{ℓ}{2} \sin θ) + M g (ℓ \sin θ) = I_{B} α_{B} ...(ii)$

From eq. (i) & (ii)

$I_{A} α_{A} = I_{B} α_{B}$

Since $I_{A} > I_{B}$ $∴$ $α_{A} < α_{B}$

Hence, $ω_{A} < ω_{B}$

Q 6 :

Match the Following [2008]

Column I Column II

(A) Potential energy of a simple pendulum (y axis) as a function of displacement (x axis) (p) [IMAGE 557]

(B) Displacement (y axis) as a function of time (x axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x-direction. (q) [IMAGE 558]

(C) Range of a projectile (y axis) as a function of its velocity (x axis) when projected at a fixed angle. (r) [IMAGE 559]

(D) The square of the time period (y axis) of a simple pendulum as a function of its length (x axis) (s) [IMAGE 560]

$A \to p; B \to q, r, s; C \to s; D \to q$
$A \to q, r, s; B \to p; C \to s; D \to q$
$A \to q, r, s; B \to p; C \to q; D \to s$
$A \to s; B \to p; C \to q; D \to q, r, s$

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(1)

(A) Potential energy is minimum at mean position and maximum at extreme position. In case of a S.H.M. we get a parabola for potential energy versus displacement graph.

(B) $S = u t$ for $a = 0$ . Therefore we get a straight line passing through the origin, as shown in graph (a).

If at $t = 0$ , $Y \neq 0$ and $Y = Y_{0}$ . Then for constant acceleration, we have graph as shown in (r).

$S = u t + \frac{1}{2} a t^{2}$ for constant positive acceleration. In this case we get a part of parabola as a graph line between $s$ versus $t$ as shown by graph (s).

(p) is ruled out because if $a$ is -ve and $v$ is positive.

$S = S_{0} + v t graph (r)$

$(C) : R = \frac{u^{2} \sin 2 θ}{g} \Rightarrow R \propto u^{2}$ [IMAGE 561]

$(D) : T = 2 π \sqrt{\frac{ℓ}{g}} \Rightarrow T^{2} \propto ℓ$ [IMAGE 562]

Q 7 :

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $ω_{1}$ and $ω_{2}$ and have total energies $E_{1}$ and $E_{2}$ , respectively. The variations of their momenta $p$ with positions $x$ are shown in the figures. If $\frac{a}{b} = n^{2}$ and $\frac{a}{R} = n,$ then the correct equation(s) is(are) [2015]

[IMAGE 563]

$E_{1} ω_{1} = E_{2} ω_{2}$
$\frac{ω_{2}}{ω_{1}} = n^{2}$
$ω_{1} ω_{2} = n^{2}$
$\frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}}$

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Select one or more options

(2, 4)

$For first harmonic oscillator,$

$Mass = m$

[IMAGE 564]--------------

$Angular frequency = ω_{1}$

$Amplitude = a$

$Total energy = E_{1}$

$Maximum momentum, p_{\max} = b$

$E_{1} = \frac{1}{2} m ω_{1}^{2} a^{2}$ ...(i)

$p_{\max} = m v_{\max} = m a ω_{1} \Rightarrow b = m a ω_{1}$

$\frac{a}{b} = \frac{1}{m ω_{1}}$ ...(ii)

$For second harmonic oscillator,$

[IMAGE 565]-----------------

$Mass = m$

$Angular frequency = ω_{2}$

$Amplitude = R$

$Maximum momentum, p_{\max} = R$

$Total energy = E_{2}$

$E_{2} = \frac{1}{2} m ω_{2}^{2} R^{2}$ ...(iii)

$p_{\max} = m v_{\max} = m ω_{2} R$

$R = m ω_{2} R \Rightarrow m ω_{2} = 1$ ...(iv)

From eqns. (ii) and (iv),

$\frac{a}{b} = \frac{ω_{2}}{ω_{1}}$ ...(v)

From eqns. (i) and (iii),

$\frac{E_{1}}{E_{2}} = \frac{ω_{1}^{2} a^{2}}{ω_{2}^{2} R^{2}}$

If $\frac{a}{b} = n^{2}$ and $\frac{a}{R} = n$ then from eqn. (v),

$\frac{ω_{2}}{ω_{1}} = n^{2}$

and from eqn. (vi),

$\frac{E_{1}}{E_{2}} = \frac{ω_{1}^{2}}{ω_{2}^{2}} \times n^{2} = \frac{ω_{1}}{ω_{2}}$

$∴$ $\frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}}$

Q 8 :

Column I Column II

(A) The object moves on the $x$ -axis under conservative force in such a way that its ''speed'' and position satisfy $v = c_{1} \sqrt{c_{2} - x^{2}}$ where $c_{1}$ and $c_{2}$ are positive constants. (p) The object executes a simple harmonic motion.

(B) The object moves on the $x$ -axis in such a way that its velocity and its displacement from the origin satisfy $v = - k x,$ where $k$ is a positive constant. (q) The object does not change its direction.

(C) The object is attached to one end of a massless spring of given spring constant. The other end of the spring is attached to the ceiling of an elevator. Initially everything is at rest. The elevator starts going upwards with a constant acceleration $a$ . The motion of the object is observed from the elevator during the period it maintains this acceleration. (r) The kinetic energy of the object keeps on decreasing.

(D) The object is projected from the earth's surface vertically upwards with a speed $2 \sqrt{\frac{G M_{e}}{R_{e}}},$ where $M_{e}$ is the mass of the earth and $R_{e}$ is the radius of the earth. Neglect forces from objects other than the earth. (s) The object can change its direction only once.

[2007]

$A \to q, r; B \to p; C \to p; D \to q, r$
$A \to p; B \to q, r; C \to p; D \to q, r$
$A \to q, r; B \to p; C \to q, r; D \to p$
$A \to p; B \to q, r; C \to q, r; D \to p$

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(2)

(A): For a simple harmonic motion $v = ω \sqrt{a^{2} - x^{2}}$ . On comparing it with $v = c_{1} \sqrt{c_{2} - x^{2}}$ this equation is SHM with $ω = c_{1}$ and $a^{2} = c_{2}$

(B): $v = - k x$

when $x$ is positive; $v$ is $- v e$ , and as $x$ decreases, $v$ decreases. Therefore kinetic energy will decrease. When $x = 0, v = 0$ . Therefore the object does not change its direction.

When $x$ is negative, $v$ is positive. But as $x$ decreases in magnitude, $v$ also decreases. Therefore kinetic energy decreases. When $x = 0, v = 0$ . Therefore the object does not change its direction.

(C): When $a = 0$ , let the spring have an extension $x$ . Then $k x = m g$

When the elevator starts going upwards with a constant acceleration, as seen by the observer in the elevator, the object is at rest.

$∴$ $m a + m g = k x'$

$\Rightarrow m a = k (x' - x)$ (Since $a$ is constant)

(D): The object is projected with a speed is $\sqrt{2}$ times the escape speed $V_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$ . Therefore the object will leave the earth. It will therefore not change the direction and keeps on moving with decreasing speed.

Q 9 :

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one dimension.

[IMAGE 566]

For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative. [2011]

Q. The phase space diagram for a ball thrown vertically up from ground is:

[IMAGE 567]
[IMAGE 568]
[IMAGE 569]
[IMAGE 570]

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(4)

When the ball is thrown upwards, at the point of throw (O) the linear momentum is in upwards direction (and has a maximum value) and the position is zero. As the time passes, the ball moves upwards and its momentum goes on decreasing and the position becomes positive. The momentum becomes zero at the topmost point.

As the time increases, the ball starts moving down with an increasing linear momentum in the downward direction (negative) and reaches back to its original position.

Q 10 :

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one dimension.

[IMAGE 571]

For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative. [2011]

Q. The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and $E_{1}$ and $E_{2}$ are the total mechanical energies respectively. Then

[IMAGE 572]

$E_{1} = \sqrt{2} E_{2}$
$E_{1} = 2 E_{2}$
$E_{1} = 4 E_{2}$
$E_{1} = 16 E_{2}$

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(3)

In SHM mechanical energy, $E \propto {(amplitude)}^{2}$

$∴$ $E_{1} \propto {(2 a)}^{2} & E_{2} \propto a^{2}$

$∴$ $\frac{E_{1}}{E_{2}} = 4$

Q 11 :

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one dimension.

[IMAGE 573]

For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative. [2011]

Q. Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

[IMAGE 574]

[IMAGE 575]
[IMAGE 576]
[IMAGE 577]
[IMAGE 578]

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(2)

When the position of the mass is at one extreme end in the positive side (the topmost point), the momentum is zero. As the mass moves towards the mean position the momentum increases in the negative direction.

As the mass is oscillating in water its amplitude will go on decreasing and the amplitude will decrease with time.

Q 12 :

When a particle of mass $m$ moves on the $x$ -axis in a potential of the form $V (x) = k x^{2}$ it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}},$ 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 $k x^{2}$ 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 $V (x) = α x^{4} (α > 0)$ for $| x |$ near the origin and becomes a constant equal to $V_{0}$ for $| x |$ $\geq X_{0}$ (see figure). [2010]

[IMAGE 579]

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

$E < 0$
$E > 0$
$V_{0} > E > 0$
$E > V_{0}$

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(3)

The particle will not perform oscillations if $E \leq 0 .$ Therefore, $E > 0 .$ If $E = V_{0},$ the potential energy will become constant as depicted in the graph given. In this case also the particle will not oscillate.

$∴$ $E > V_{0}$

$∴$ $V_{0} > E > 0$

Q 13 :

When a particle of mass $m$ moves on the $x$ -axis in a potential of the form $V (x) = k x^{2}$ it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}},$ 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 $k x^{2}$ 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 $V (x) = α x^{4} (α > 0)$ for $| x |$ near the origin and becomes a constant equal to $V_{0}$ for $| x |$ $\geq X_{0}$ (see figure). [2010]

[IMAGE 580]

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

$A \sqrt{\frac{m}{α}}$
$\frac{1}{A} \sqrt{\frac{m}{α}}$
$A \sqrt{\frac{α}{m}}$
$\frac{1}{A} \sqrt{\frac{α}{m}}$

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(2)

Potential energy, $V \propto x^{4}$ given

$∴$ $α = \frac{Potential energy}{x^{4}} = \frac{M L^{2} T^{- 2}}{L^{4}} = [M L^{- 2} T^{- 2}]$

Now, $\frac{1}{A} \sqrt{\frac{m}{α}} = \frac{1}{L} \sqrt{\frac{M}{M L^{- 2} T^{- 2}}} = T$

Q 14 :

When a particle of mass $m$ moves on the $x$ -axis in a potential of the form $V (x) = k x^{2}$ it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}},$ 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 $k x^{2}$ 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 $V (x) = α x^{4} (α > 0)$ for $| x |$ near the origin and becomes a constant equal to $V_{0}$ for $| x |$ $\geq X_{0}$ (see figure). [2010]

[IMAGE 581]

Q. The acceleration of this particle for $| x |$ $> X_{0}$ is

proportional to $V_{0}$
proportional to $\frac{V_{0}}{m X_{0}}$
proportional to $\sqrt{\frac{V_{0}}{m X_{0}}}$
zero

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(4)

$F = \frac{- d V (x)}{d x}$

$∵$ $As V (x) = constant for x > X_{0}$

$∴$ $F = 0 and hence a = 0 for x > X_{0}$

Topic Question Set

Q 1 :

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5 s. In another 10 s it will decrease to $α$ times its original magnitude, where $α$ equals [2013]

Q 2 :

A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector $\vec{a}$ is correctly shown in: [2002]

Q 3 :

A particle executes simple harmonic motion between $x = - A$ and $x = + A .$ The time taken for it to go from 0 to $\frac{A}{2}$ is $T_{1}$ and to go from $\frac{A}{2}$ to $A$ is $T_{2}$ . Then [2001]

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	Column I		Column II
(A)	Potential energy of a simple pendulum (y axis) as a function of displacement (x axis)	(p)	[IMAGE 557]
(B)	Displacement (y axis) as a function of time (x axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x-direction.	(q)	[IMAGE 558]
(C)	Range of a projectile (y axis) as a function of its velocity (x axis) when projected at a fixed angle.	(r)	[IMAGE 559]
(D)	The square of the time period (y axis) of a simple pendulum as a function of its length (x axis)	(s)	[IMAGE 560]

	Column I		Column II
(A)	The object moves on the $x$ -axis under conservative force in such a way that its ''speed'' and position satisfy $v = c_{1} \sqrt{c_{2} - x^{2}}$ where $c_{1}$ and $c_{2}$ are positive constants.	(p)	The object executes a simple harmonic motion.
(B)	The object moves on the $x$ -axis in such a way that its velocity and its displacement from the origin satisfy $v = - k x,$ where $k$ is a positive constant.	(q)	The object does not change its direction.
(C)	The object is attached to one end of a massless spring of given spring constant. The other end of the spring is attached to the ceiling of an elevator. Initially everything is at rest. The elevator starts going upwards with a constant acceleration $a$ . The motion of the object is observed from the elevator during the period it maintains this acceleration.	(r)	The kinetic energy of the object keeps on decreasing.
(D)	The object is projected from the earth's surface vertically upwards with a speed $2 \sqrt{\frac{G M_{e}}{R_{e}}},$ where $M_{e}$ is the mass of the earth and $R_{e}$ is the radius of the earth. Neglect forces from objects other than the earth.	(s)	The object can change its direction only once.

Topic Question Set

Q 1 : The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5 s. In another 10 s it will decrease to α times its original magnitude, where α equals [2013]

Q 2 : A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector a→ is correctly shown in: [2002]

Q 3 : A particle executes simple harmonic motion between x=-A and x=+A. The time taken for it to go from 0 to A2 is T1 and to go from A2 to A is T2. Then [2001]

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Q 1 :

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5 s. In another 10 s it will decrease to $α$ times its original magnitude, where $α$ equals [2013]

Q 2 :

A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector $\vec{a}$ is correctly shown in: [2002]

Q 3 :

A particle executes simple harmonic motion between $x = - A$ and $x = + A .$ The time taken for it to go from 0 to $\frac{A}{2}$ is $T_{1}$ and to go from $\frac{A}{2}$ to $A$ is $T_{2}$ . Then [2001]