JEE Advanced PYQ Time Period, Frequency, Simple Pendulum and Spring Pendulum

Q 1 :

As shown in the figures, a uniform rod OO' of length $l$ is hinged at the point O and held in place vertically between two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end (O') of the rod, as shown in Fig. 1 and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $f_{1}$ . On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $f_{2}$ . Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_{1}}{f_{2}}$ is: [2025]

[IMAGE 528]

$2$
$\sqrt{2}$
$\sqrt{\frac{5}{2}}$
$\sqrt{\frac{2}{5}}$

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(3)

[IMAGE 529]

$τ_{net} = K (θ ℓ) ℓ + K (θ . \frac{ℓ}{2}) \frac{ℓ}{2}$

$τ_{net} = (K θ \frac{ℓ}{2}) \frac{ℓ}{2} \times 2$

$= \frac{5 K ℓ^{2}}{4} θ = ℓ α = \frac{K ℓ^{2}}{2} θ = ℓ α$

As $α = ω^{2} θ$

So, $\frac{ω_{1}^{2}}{ω_{2}^{2}} = \frac{5 K ℓ^{2}}{\frac{4 K ℓ^{2}}{2}} = \frac{5}{2}$ or, $\frac{ω_{1}}{ω_{2}} = \sqrt{\frac{5}{2}}$

Q 2 :

A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at $t = 0$ . It then executes simple harmonic motion with angular frequency $ω = \frac{π}{3} rad/s .$ Simultaneously at $t = 0$ , a small pebble is projected with speed $v$ from point $P$ at an angle of $45 °$ as shown in the figure. Point $P$ is at a horizontal distance of 10 m from O. If the pebble hits the block at $t = 1 s$ , the value of $v$ is (take $g = 10 {m/s}^{2}$ ) [2012]

[IMAGE 530]

$\sqrt{50} m/s$
$\sqrt{51} m/s$
$\sqrt{52} m/s$
$\sqrt{53} m/s$

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(1)

Here, we don't need to consider the SHM part rather we will focus on projectile part only. Since the pebble hits the block after 1 sec. We can easily calculate the speed of projection (V) from this time of flight.

Time of flight of projectile,

$T = \frac{2 V \sin θ}{g}$

$∴$ $1 = \frac{2 V \sin 45 °}{g}$

$∴$ $V = \sqrt{50} {ms}^{- 1}$

Hence pebble is projected with a speed $V = \sqrt{50} {ms}^{- 1}$

Q 3 :

The center of a disk of radius $r$ and mass $m$ is attached to a spring of spring constant $k$ , inside a ring of radius $R > r$ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disk, the time period of oscillation of center of mass of the disk is written as $T = \frac{2 π}{ω} .$ The correct expression for $ω$ is ( $g$ is the acceleration due to gravity): [2025]

[IMAGE 531]

$\sqrt{\frac{2}{3} (\frac{g}{R - r} + \frac{k}{m})}$
$\sqrt{\frac{2 g}{3 (R - r)} + \frac{k}{m}}$
$\sqrt{\frac{1}{6} (\frac{g}{R - r} + \frac{k}{m})}$
$\sqrt{\frac{1}{4} (\frac{g}{R - r} + \frac{k}{m})}$

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(1)

$From figure, total energy$

$E = \frac{1}{2} k {(R - r)}^{2} θ^{2} + m g (R - r) (1 - \cos θ) + \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{m r^{2}}{2}) ω^{2}$

$Differentiating w.r.t. t,$

$0 = \frac{1}{2} k {(R - r)}^{2} \cdot 2 θ \frac{d θ}{d t} + m g (R - r) \frac{d}{d t} (2 \frac{θ^{2}}{4}) + \frac{1}{2} m \cdot 2 v \frac{d v}{d t} + \frac{m r^{2}}{4} \cdot 2 ω \frac{d ω}{d t}$

$\Rightarrow 0 = k {(R - r)}^{2} θ \frac{d θ}{d t} + m g (R - r) θ \frac{d θ}{d t} + m v \frac{d v}{d t} + \frac{m r^{2}}{2} ω \frac{d ω}{d t}$

[IMAGE 532]

Also, $\frac{d θ}{d t} = \frac{v}{R - r} \Rightarrow \frac{d^{2} θ}{d t^{2}} = \frac{1}{R - r} \frac{d v}{d t} = \frac{a}{R - r}$

$∴$ $k {(R - r)}^{2} θ \frac{v}{R - r} + m g (R - r) θ \frac{v}{R - r} = - m v α r - \frac{m r^{2}}{2} \frac{v}{r} α$

$\Rightarrow k (R - r) + m g θ = - \frac{3}{2} m r α$

$\Rightarrow - [k (R - r) + m g] θ = \frac{3}{2} m (R - r) \frac{d^{2} θ}{d t^{2}}$

$\Rightarrow - \frac{2}{3} [\frac{k}{m} + \frac{g}{R - r}] = \frac{d^{2} θ}{d t^{2}}$

Now comparing this equation with standard equation of SHM, we get

$ω = \sqrt{\frac{2}{3} [\frac{k}{m} + \frac{g}{R - r}]}$

Q 4 :

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is $V_{0}$ and its pressure is $P_{0}$ . The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency [2013]

$\frac{1}{2 π} \frac{A γ P_{0}}{V_{0} M}$
$\frac{1}{2 π} \frac{V_{0} M P_{0}}{A^{2} γ}$
$\frac{1}{2 π} \sqrt{\frac{A^{2} γ P_{0}}{M V_{0}}}$
$\frac{1}{2 π} \sqrt{\frac{M V_{0}}{A γ P_{0}}}$

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(3)

[IMAGE 533]

$\frac{M g}{A} = P_{0}$

$P_{0} V_{0}^{γ} = P V^{γ}$

$M g = P_{0} A . . . (1)$

Let piston is displaced by distance $x$

$P_{0} A x_{0}^{γ} = P A {(x_{0} - x)}^{γ}$

$P = \frac{P_{0} x_{0}^{γ}}{{(x_{0} - x)}^{γ}}$

$M g - (\frac{P_{0} x_{0}^{γ}}{(x_{0} - x)^{γ}}) A = F_{restoring}$

$P_{0} A (1 - \frac{x_{0}^{γ}}{{(x_{0} - x)}^{γ}}) = F_{restoring}$ $[x_{0} - x \approx x_{0}]$

$P_{0} A (1 - \frac{1}{{(1 - \frac{x}{x_{0}})}^{γ}}) = P_{0} A [1 - {(1 - \frac{x}{x_{0}})}^{γ}]$

$= P_{0} A [1 - (1 + \frac{γ x}{x_{0}})]$

$F = - \frac{γ P_{0} A x}{x_{0}}$ $\Rightarrow M ω^{2} x = \frac{γ P_{0} A x}{x_{0}}$ $\Rightarrow ω = \sqrt{\frac{γ P_{0} A}{x_{0} M}}$

$∴$ $Frequency with which piston executes SHM$

$f = \frac{1}{2 π} \sqrt{\frac{γ P_{0} A}{x_{0} M}}$ $= \frac{1}{2 π} \sqrt{\frac{γ P_{0} A^{2}}{M V_{0}}}$

Q 5 :

The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is [2009]

[IMAGE 534]

$\frac{k_{1} A}{k_{2}}$
$\frac{k_{2} A}{k_{1}}$
$\frac{k_{1} A}{k_{1} + k_{2}}$
$\frac{k_{2} A}{k_{1} + k_{2}}$

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(4)

If the spring of spring constant $k_{1}$ is compressed by $x_{1}$ and that of spring constant $k_{2}$ is compressed by $x_{2}$ , then

$x_{1} + x_{2} = A . . . (i)$

and $k_{1} x_{1} = k_{2} x_{2}$ $\Rightarrow x_{2} = \frac{k_{1} x_{1}}{k_{2}} . . . (i i)$

Solving eqs. (i) & (ii) we get

$x_{1} = \frac{k_{2} A}{k_{2} + k_{1}}$

Q 6 :

A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upward according to the relation $y = K t^{2}, (K = 1 {m/s}^{2})$ where $y$ is the vertical displacement. The time period now becomes $T_{2}$ . The ratio $\frac{T_{1}^{2}}{T_{2}^{2}}$ is $(g = 10 {m/s}^{2})$ [2005]

5/6
6/5
1
4/5

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(2)

$Given : y = k t^{2}$

$∴$ $\frac{d y}{d t} = v = 2 k t$ and, $\frac{d^{2} y}{d t^{2}} = a = 2 k$

or $a_{y} = 2 {m/s}^{2}$ $(∵ k = 1 {m/s}^{2} given)$

We know that $T = 2 π \sqrt{\frac{l}{g}}$ or, $T_{1} = 2 π \sqrt{\frac{l}{g}}$

$T_{2} = 2 π \sqrt{\frac{l}{g + a_{y}}}$

$∴$ $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{g + a_{y}}{g}$ $\Rightarrow \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{12}{10} = \frac{6}{5}$

Q 7 :

A block P of mass $m$ is placed on a horizontal frictionless plane. A second block of same mass $m$ is placed on it and is connected to a spring of spring constant $k$ , the two blocks are pulled by distance $A$ . Block $Q$ oscillates without slipping. What is the maximum value of frictional force between the two blocks? [2004]

[IMAGE 535]

$\frac{k A}{2}$
$k A$
$μ_{s} m g$
$zero$

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(1)

Here acceleration is same for both $P$ and $Q$ as block $Q$ oscillates but does not slip. The $(P - Q)$ system oscillates with angular frequency $ω$ . The spring is stretched by $A$ .

Angular frequency of the system,

$ω = \sqrt{\frac{k}{m + m}} = \sqrt{\frac{k}{2 m}}$

Maximum acceleration of the system in SHM

$a_{\max} = A ω^{2} = A {(\sqrt{\frac{k}{2 m}})}^{2} = \frac{k A}{2 m}$

This acceleration to the lower block is provided by friction

$∴$ Maximum force of friction

$f_{\max} = m a_{\max} = m (\frac{k A}{2 m}) = \frac{k A}{2}$

Q 8 :

Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod of length 10 cm. This system is suspended vertically from a rigid ceiling using a thin wire attached to its centre of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is $1.2 \times 10^{- 8} {N m rad}^{- 1} .$ The angular frequency of the oscillations is $n \times 10^{- 3} {rad s}^{- 1} .$ The value of $n$ is ______. [2023]

[IMAGE 536]

(10)

[IMAGE 537]

$Time period of torsional pendulum$

$T = \frac{2 π}{ω} = 2 π \sqrt{\frac{I}{C}}$

$∴$ $ω = \sqrt{\frac{C}{I}}$

Where $I =$ moment of inertia

$I = (30) {(4)}^{2} + (20) {(6)}^{2} = 1200 {g cm}^{2}$

or, $I = 1.2 \times 10^{- 4} {kg m}^{2}$ $= \sqrt{\frac{C}{I}} ω = \sqrt{\frac{1.2 \times 10^{- 8}}{1.2 \times 10^{- 4}}}$

$[∵ Torsional constant of the wire C = 1.2 \times 10^{- 8} {N m rad}^{- 1} given]$

$∴$ $ω = \sqrt{10^{- 4}} = 10^{- 2}$

$∴$ $n \times 10^{- 3} = 10^{- 2}$

$∴$ $n = 10$

Q 9 :

On a frictionless horizontal plane, a bob of mass $m = 0.1 kg$ is attached to a spring with natural length $l_{0} = 0.1 m .$ The spring constant is $k_{1} = 0.009 {Nm}^{- 1}$ when the length of the spring $l > l_{0}$ and is $k_{2} = 0.016 {N m}^{- 1}$ when $l < l_{0} .$ Initially the bob is released from $l = 0.15 m .$ Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T = (n π) s,$ then the integer closest to $n$ is _______. [2022]

(6)

$We have$

$ℓ > ℓ_{0} \Rightarrow k = k_{1}; ℓ < ℓ_{0} \Rightarrow k = k_{2}$

Time period of oscillation,

$T = π \sqrt{\frac{m}{k_{1}}} + π \sqrt{\frac{m}{k_{2}}}$ $= π \sqrt{\frac{0.1}{0.009}} + π \sqrt{\frac{0.1}{0.016}}$

$= \frac{π}{0.3} + \frac{π}{0.4}$ $= π \times \frac{0.70}{0.12}$ $= 5.83 π \approx 6 π$

So, $n = 6$

Q 10 :

A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0 {Nm}^{- 1}$ and the mass of the block is $2.0 kg .$ Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass $1.0 kg$ moving with a speed of $2.0 {ms}^{- 1}$ collides elastically with the first block. The collision is such that the $2.0 kg$ block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is ________. [2018]

[IMAGE 538]

(2.09)

Let velocities of $1 kg$ and $2 kg$ blocks just after collision be $v_{1}$ and $v_{2}$ respectively.

Just after collision

[IMAGE 539]

From momentum conservation principle,

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$0 + 1 \times 2 = - 1 v_{1} + 2 v_{2} ..... (i)$

Collision is elastic. Hence $e = 1$

$e = 1 = \frac{2 - 0}{v_{1} + v_{2}}$

$\Rightarrow v_{2} + v_{1} = 2 ..... (ii)$

From eqs. (i) and (ii),

$v_{2} = \frac{4}{3} m/s, v_{1} = \frac{2}{3} m/s$

$2 kg$ block will perform SHM after collision, so spring returns to its unstretched position for the first time after,

$t = \frac{T}{2} = π \sqrt{\frac{m}{k}} = π \sqrt{\frac{2}{2}} = π = 3.14 s$

Distance or required separation between the blocks

$= | v_{1} | t = \frac{2}{3} \times 3.14 = 2.093 = 2.09 m$

Q 11 :

A block with mass M is connected by a massless spring with stiffness constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position $x_{0}$ . Consider two cases: (i) when the block is at $x_{0}$ , and (ii) when the block is at $x = x_{0} + A .$ In both the cases, a particle with mass $m (m < M)$ is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass $m$ is placed on the mass $M$ ? [2016]

The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m + M}},$ whereas in the second case it remains unchanged.
The final time period of oscillation in both the cases is same.
The total energy decreases in both the cases.
The instantaneous speed at $x_{0}$ of the combined masses decreases in both the cases.

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Select one or more options

(1, 2, 4)

Case (i) : Applying principle of conservation of linear momentum.

[IMAGE 540]

$M V_{1} = (M + m) V_{2} .... (i)$

$M (A_{1} \times ω_{1}) = (M + m) (A_{2} \times ω_{2})$

$∴$ $M A_{1} \times \sqrt{\frac{K}{M}} = (M + m) A_{2} \times \sqrt{\frac{K}{M + m}}$

$∴$ $A_{2} = \sqrt{\frac{M}{M + m}} A_{1} \Rightarrow \frac{A_{2}}{A_{1}} = \sqrt{\frac{M}{M + m}}$

Also, $E_{1} = \frac{1}{2} M V_{1}^{2}$

and $E_{2} = \frac{1}{2} (M + m) V_{2}^{2}$ $(∵ V_{2} = (\frac{M}{M + m}) V_{1} from eq (i))$

$\Rightarrow E_{2} = \frac{1}{2} (M + m) {(\frac{M}{M + m})}^{2} V_{1}^{2}$

$= \frac{1}{2} (\frac{M}{M + m}) V_{1}^{2}$

Clearly, $E_{1} > E_{2}$

The new time period $T_{2} = 2 \sqrt{\frac{m + M}{K}}$

Instantaneous speed at $X_{0}$ of the combined masses $V_{2} = \frac{M V_{1}}{M + m} < V_{1}$

Case (ii) : The new time period $T_{2} = 2 \sqrt{\frac{m + M}{K}}$

Also, $A_{2} = A_{1} and E_{2} = E_{1}$

In this case also, instantaneous speed at $X_{0}$ of the combined masses decreases.

Q 12 :

A particle of mass $m$ is attached to one end of a massless spring of force constant $k$ , lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t = 0$ with an initial velocity $u_{0}$ . When the speed of the particle is $0.5 u_{0},$ it collides elastically with a rigid wall. After this collision [2013]

The speed of the particle when it returns to its equilibrium position is $u_{0}$ .
The time at which the particle passes through the equilibrium position for the first time is $t = π \sqrt{\frac{m}{k}} .$
The time at which the maximum compression of the spring occurs is $t = \frac{4 π}{3} \sqrt{\frac{m}{k}} .$
The time at which the particle passes through the equilibrium position for the second time is $t = \frac{5 π}{3} \sqrt{\frac{m}{k}} .$

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Select one or more options

(1, 4)

The particle collides elastically with rigid wall.

$∴$ $e = \frac{V}{0.5 u_{0}} = 1 \Rightarrow V = 0.5 u_{0}$

i.e., the particle rebounds with the same speed. Therefore the particle will return to its equilibrium position with speed $u_{0}$ . So, (1) is correct.

[IMAGE 541]

The velocity of the particle becomes $0.5 u_{0}$ after time $t$ .

Using equation $V = V_{\max} \cos ω t$

$0.5 u_{0} = u_{0} \cos ω t$

$∴$ $\frac{π}{3} = \frac{2 π}{T} \times t \Rightarrow t = \frac{T}{6}$

The time period $T = 2 π \sqrt{\frac{m}{k}}$ $∴$ $t = \frac{π}{3} \sqrt{\frac{m}{k}}$

The time taken by the particle to pass through the equilibrium for the first time $= 2 t = \frac{2 π}{3} \sqrt{\frac{m}{k}}$

So, (2) is wrong.

The time taken for the maximum compression $= t_{A B} + t_{B A} + t_{A C}$

$= \frac{π}{3} \sqrt{\frac{m}{k}} + \frac{π}{3} \sqrt{\frac{m}{k}} + \frac{π}{2} \sqrt{\frac{m}{k}}$

$= π \sqrt{\frac{m}{k}} [\frac{1}{3} + \frac{1}{3} + \frac{1}{2}]$

$= \frac{7 π}{6} \sqrt{\frac{m}{k}}$

So, (3) is wrong.

The time taken for particle to pass through the equilibrium position second time

$= 2 [\frac{π}{3} \sqrt{\frac{m}{k}}] + π \sqrt{\frac{m}{k}}$

$= π \sqrt{\frac{m}{k}} (\frac{2}{3} + 1)$

$= \frac{5 π}{3} \sqrt{\frac{m}{k}}$

So (4) is correct.

Q 13 :

List I describes four systems, each with two particles A and B in relative motion, as shown in figures. List II gives possible magnitudes of their relative velocities (in ${ms}^{- 1}$ ) at time $t = \frac{π}{3} s .$ [2022]

List-I List-II

(I) A and B are moving on a horizontal circle of radius 1 m with uniform angular speed $ω = 1 {rad s}^{- 1} .$ The initial angular positions of A and B at time $t = 0$ are $θ = 0$ and $θ = \frac{π}{2},$ respectively.
[IMAGE 542] (P) $\frac{\sqrt{3} + 1}{2}$

(II) Projectiles A and B are fired (in the same vertical plane) at $t = 0$ and $t = 0.1 s$ respectively, with the same speed $v = \frac{5 π}{\sqrt{2}} {m s}^{- 1}$ and at $45 °$ from the horizontal plane. The initial separation between A and B is large enough so that they do not collide. $(g = 10 {m s}^{- 2})$
[IMAGE 543] (Q) $\frac{\sqrt{3} - 1}{\sqrt{2}}$

(III) Two harmonic oscillators A and B moving in the $x$ direction according to
$x_{A} = x_{0} \sin (\frac{t}{t_{0}})$ and $x_{B} = x_{0} \sin (\frac{t}{t_{0}} + \frac{π}{2}),$ respectively, starting at $t = 0$ . Take $x_{0} = 1 m, t_{0} = 1 s .$
[IMAGE 544] (R) $\sqrt{10}$

(IV) Particle A is rotating in a horizontal circular path of radius 1 m on the $x y$ plane, with constant angular speed $ω = 1 {rad s}^{- 1} .$ Particle B is moving up at a constant speed $3 {m s}^{- 1}$ in the vertical direction as shown in the figure. (Ignore gravity.)
[IMAGE 545] (S) $\sqrt{2}$

(T) $\sqrt{25 π^{2} + 1}$

Which one of the following options is correct?

(I) → (R); (II) → (T); (III) → (P); (IV) → (S)
(I) → (S); (II) → (P); (III) → (Q); (IV) → (R)
(I) → (S); (II) → (T); (III) → (P); (IV) → (R)
(I) → (T); (II) → (P); (III) → (R); (IV) → (S)

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(3)

$(I)$ $V_{B A}^{2} = V_{B}^{2} + V_{A}^{2} - 2 V_{B} V_{A} \cos θ$

As $ω_{A} = ω_{B}$ , $θ = 90 °$ remains constant.

Also, $V_{A} = V_{B} = 1 m/s$ $[∵ V = ω R]$

$V_{B A} = \sqrt{2} m/s$

So $I \to S .$

$(II)$ ${\vec{u}}_{A} = \frac{5 π}{2} \hat{i} + \frac{5 π}{2} \hat{j}$

${\vec{V}}_{A} |_{t = 0.1 sec} = \frac{5 π}{2} \hat{i} + (\frac{5 π}{2} - 10 \times 0.1) \hat{j} = \frac{5 π}{2} \hat{i} + (\frac{5 π}{2} - 1) \hat{j}$

${\vec{V}}_{B} |_{t = 0.1 sec} = - \frac{5 π}{2} \hat{i} + \frac{5 π}{2} \hat{j}$

After $t = 0.1$ sec, both projectile came in air. So the relative acceleration is zero. So relative velocity should not change after it.

$V_{rel} = V_{rel} (t = 0.1 sec) =$ $| 5 π \hat{i} - \hat{j} |$ $= \sqrt{25 π^{2} + 1}$

So $II \to T .$

$(III)$ $x = x_{A} - x_{B}$

$= x_{0} \sin t - x_{0} \sin (t + \frac{π}{2})$ $[∵ t_{0} = 1]$

$= \sin t - \cos t = \sqrt{2} (\frac{1}{\sqrt{2}} \sin t - \frac{1}{\sqrt{2}} \cos t)$

$= \sqrt{2} \sin (t - \frac{π}{4})$

$V_{rel} = \frac{d x}{d t} = \sqrt{2} \cos (t - \frac{π}{4})$ $= \sqrt{2} \cos (\frac{π}{3} - \frac{π}{4})$

$= \sqrt{2} \times \frac{\sqrt{3} + 1}{2 \sqrt{2}} = \frac{\sqrt{3} + 1}{2}$

So, $III \to P .$

$(IV)$ ${\vec{V}}_{A} and {\vec{V}}_{B} are always perpendicular$

So, $| {\vec{V}}_{B A} |$ $= \sqrt{V_{A}^{2} + V_{B}^{2}} = \sqrt{3^{2} + 1^{2}} = \sqrt{10} m/s$

So $IV \to R .$

Q 14 :

Two particles, 1 and 2, each of mass $m$ , are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_{0}$ , are oscillating with amplitude $a$ and angular frequency $ω$ . Thus, their positions at time $t$ are given by $x_{1} (t) = (x_{0} + d) + a \sin ω t$ and $x_{2} (t) = (x_{0} - d) - a \sin ω t,$ respectively, where $d > 2 a .$ Particle 3 of mass $m$ moves towards this system with speed $u_{0} = \frac{a ω}{2},$ and undergoes instantaneous elastic collision with particle 2, at time $t_{0}$ . Finally, particles 1 and 2 acquire a center of mass speed $v_{cm}$ and oscillate with amplitude $b$ and the same angular frequency $ω$ . [2024]

[IMAGE 546]

Q. If the collision occurs at time $t_{0} = 0,$ the value of $\frac{v_{cm}}{(a ω)}$ will be ________. [2024]

(0.75)

$At time t_{0} = 0 collision occurs$

$Before collision$

[IMAGE 547]

$V_{C M} = \frac{m (\frac{a ω}{2}) + m (a ω)}{m + m} = \frac{3 a ω}{4}$

$∴$ $\frac{V_{C M}}{a ω} = \frac{3}{4} = 0.75$

Q 15 :

Two particles, 1 and 2, each of mass $m$ , are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_{0}$ , are oscillating with amplitude $a$ and angular frequency $ω$ . Thus, their positions at time $t$ are given by $x_{1} (t) = (x_{0} + d) + a \sin ω t$ and $x_{2} (t) = (x_{0} - d) - a \sin ω t,$ respectively, where $d > 2 a .$ Particle 3 of mass $m$ moves towards this system with speed $u_{0} = \frac{a ω}{2},$ and undergoes instantaneous elastic collision with particle 2, at time $t_{0}$ . Finally, particles 1 and 2 acquire a center of mass speed $v_{cm}$ and oscillate with amplitude $b$ and the same angular frequency $ω$ .

[IMAGE 548]

Q. If the collision occurs at time $t_{0} = \frac{π}{2 ω},$ then the value of $\frac{4 b^{2}}{a^{2}}$ will be _______. [2024]

(4.25)

$If the collision occurs at time$ $t_{0} = \frac{π}{2 ω} = \frac{T}{4}$

Particles are at extreme position.

[IMAGE 549]

Using work-energy theorem, $W_{spring} = Δ K$

$\frac{1}{2} k {(2 b)}^{2} - \frac{1}{2} k {(2 a)}^{2} = 2 \times \frac{1}{2} m {(\frac{a ω}{4})}^{2}$

$4 k b^{2} - 4 k a^{2} = 2 \times m \times \frac{a^{2}}{16} \times \frac{2 k}{m}$

$\Rightarrow 4 b^{2} = \frac{17}{4} a^{2}$

$∴$ $\frac{4 b^{2}}{a^{2}} = 4.25$

Topic Question Set

Q 5 :

The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is [2009]

[IMAGE 534]

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	List-I		List-II
(I)	A and B are moving on a horizontal circle of radius 1 m with uniform angular speed $ω = 1 {rad s}^{- 1} .$ The initial angular positions of A and B at time $t = 0$ are $θ = 0$ and $θ = \frac{π}{2},$ respectively. [IMAGE 542]	(P)	$\frac{\sqrt{3} + 1}{2}$
(II)	Projectiles A and B are fired (in the same vertical plane) at $t = 0$ and $t = 0.1 s$ respectively, with the same speed $v = \frac{5 π}{\sqrt{2}} {m s}^{- 1}$ and at $45 °$ from the horizontal plane. The initial separation between A and B is large enough so that they do not collide. $(g = 10 {m s}^{- 2})$ [IMAGE 543]	(Q)	$\frac{\sqrt{3} - 1}{\sqrt{2}}$
(III)	Two harmonic oscillators A and B moving in the $x$ direction according to $x_{A} = x_{0} \sin (\frac{t}{t_{0}})$ and $x_{B} = x_{0} \sin (\frac{t}{t_{0}} + \frac{π}{2}),$ respectively, starting at $t = 0$ . Take $x_{0} = 1 m, t_{0} = 1 s .$ [IMAGE 544]	(R)	$\sqrt{10}$
(IV)	Particle A is rotating in a horizontal circular path of radius 1 m on the $x y$ plane, with constant angular speed $ω = 1 {rad s}^{- 1} .$ Particle B is moving up at a constant speed $3 {m s}^{- 1}$ in the vertical direction as shown in the figure. (Ignore gravity.) [IMAGE 545]	(S)	$\sqrt{2}$
		(T)	$\sqrt{25 π^{2} + 1}$

Topic Question Set

Q 5 : The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is [2009] [IMAGE 534]

Q 6 : A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y=Kt2, (K=1 m/s2) where y is the vertical displacement. The time period now becomes T2. The ratio T12T22 is (g=10 m/s2) [2005]

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Q 5 :

The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is [2009]

[IMAGE 534]