Oscillations Test

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Oscillations Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

A block of mass $$200$$ g executing SHM under the influence of a spring of spring constant $$k = 90 N m^{-1}$$ and a damping constant $$b = 40 g s^{-1}$$. Time taken for its amplitude of vibrations to drop to half of its initial values (Given, In $$(1/2) = -0.693)$$

A
$$7$$s

B
$$9$$s

C
$$4$$s

D
$$11$$s

Solution

Given data,
mass $$m=200g$$
Spring constant $$k=90Nm^{-1}$$
Damping constant $$b=40gs^{-1}$$
To calculate: Time taken for the amplitude of vibration to drop to half of the initial value
We know that amplitude at any time t can be given as:
$$A(t)=A_0e^{-\dfrac{bt}{2m}}$$

or $$T_{1/2}=\dfrac{-0.693l×2×0.2}{40×10^{−3}}=6.93s$$
Time taken for its amplitude of vibrations to drop to half of its initial values is $$7s$$
Question 2
1 / -0

A thin fixed ring of radius 1m has a positive charge $$1 \times 10^{-5}$$ C uniformly distributed over it. A particle of mass 0.9 g and having a negative charge of $$1 \times 10^-6$$ C is placed on the axis at a distance of 1 cm from the centre of the ring. Calculate the time period of oscillations.

A
$$0.5 secs$$

B
$$9.28 secs$$

C
$$0.628 secs$$

D
$$0.1 secs$$

Solution

The negatively charged particle kept on the axis will get attracted towards the positively charged ring. As it approaches the center, the inertia of the particle will make the charge to move to the other side of the ring. Simultaneously, the attractive force due to the ring will prevent it from going to infinity. Thus, it goes to one more extreme and returns back. Thus the charge starts oscillation about the center of the ring.
The electric field at a distance x from the centre on the axis of a ring, distanct x <<R is given by $$E=\dfrac{kQx}{R^3}$$. Net force on negatively charged particle would be qE and towards the centre of the ring. Hence, we can write $$F=-\dfrac{kQqx}{R^3}$$
Acceleration of the charge is $$a=F/m=-\dfrac{kQqx}{mR^3}=-w^2 x$$.
Time period of oscillations is $$T=2/w=2(mR^3/kQq)$$. Substituting the values , we get
T=0.628 s
Question 3
1 / -0

A thin rod of length Land uniform area of cross-section S is pivoted at its lowest point P inside a stationary, homogenous and non-viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through P. The density $$d_1$$ of the material of the rod is smaller than the density $$d_2$$ of the liquid. The rod is displaced by a small angle $$\theta$$ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters.

A
$$\sqrt{\dfrac{3}{3} \left ( \dfrac{d_2 \, - \, d_1}{d_1} \right ) \dfrac{g}{L}}$$

B
$$\sqrt{\dfrac{3}{4} \left ( \dfrac{d_2 \, - \, d_1}{d_1} \right ) \dfrac{g}{L}}$$

C
$$\sqrt{\dfrac{4}{2} \left ( \dfrac{d_2 \, - \, d_1}{d_1} \right ) \dfrac{g}{L}}$$

D
$$\sqrt{\dfrac{3}{2} \left ( \dfrac{d_2 \, - \, d_1}{d_1} \right ) \dfrac{g}{L}}$$

Solution
Question 4
1 / -0

A simple harmonic oscillator of angular frequency 2 rad $$s^{-1}$$ is acted upon by an external force $$F= sin tN.$$ If the oscillator is at rest in its equilibrium position at t=0, its position at later times is proportional to

A
$$ sin t+ \frac{1}{2}cos 2t$$

B
$$ cos t -\frac{1}{2}sin2t$$

C
$$sint -\frac{1}{2}sin2t$$

D
$$sin t +\frac{1}{2}sin 2t$$

Solution

The solution contains term proportional to $$sin t$$, $$sin 2t$$ & the only option consistent with initial conditions is (C)
Question 5
1 / -0

What kind of combination of springs does this arrangement in figure shown belong to

A
Series

B
Parallel

C
Hybrid (Series and parallel)

D
None of the above

Solution

when a body is pulled lower through small displacement y. lower spring gets compressed by y, while upper spring elongate by y. hence restoring forces $$F_1$$ and $$F_2$$ set up in both these springs will act in the same direction.

Net restoring force will be $$F = F_1 + F_2 \implies F = - k_1y -k_2y = - (k_1 + k_2) y $$

If k is the is equivalent spring constant then F = -ky thus $$k = k_1 + k_2$$. This happens only when springs are connected in parallel

The correct option is (b)
Question 6
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the system shown in $$Fig. 6.330$$ is in equilibrium. Masses $${ m }_{ 1 }$$ and $${ m }_{ 2 }$$ are $$2 kg$$ and $$8 kg$$, respectively. Spring constants $${ k }_{ 1 }$$ and $${ k }_{ 1 }$$ are $$50 N{ m }^{ -1 }$$ and $$70 N{ m }^{ -1 }$$, respectively. If the compression in second spring is $$0.5 m$$. What is the compression in first spring? (Both springs have natural length initially.)

A
$$1.3 m$$

B
$$-0.5 m$$

C
$$0.5 m$$

D
$$0.9 m$$

Solution
Question 7
1 / -0

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 $$\omega =\pi /3$$ rad/s. Simultaneously at $$t=0$$, a small pebble is projected with speed v form point P at an angle of $$45^o$$ as shown in the figure. Point P is at a horizontal distance of $$10m$$ from O. If the pebble hits the block at $$t=1$$s, the value of v is (take $$g=10m/s^2$$)

A
$$\sqrt{50}$$ m/s

B
$$\sqrt{51}$$ m/s

C
$$\sqrt{52}$$ m/s

D
$$\sqrt{53}$$ m/s

Solution

The block starts its oscillation from its positive extremum.
$$\therefore x_{block}(t)=4.9+0.2\cos \omega t$$
$$\therefore x_{block}(t=1s)=4.9+0.2\cos \left (\cfrac {\pi}{3}\times 1\right)=5$$ $$m$$
$$\therefore$$ Range of pebble $$=5m$$
So, $$\cfrac {v^2\sin 2\theta}{g}=5$$
$$\implies v=\sqrt {50}$$ $$m/s$$
Question 8
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A block of mass $$m$$ has initial velocity $$u$$ having direction towards '+x axis'. The block stops after covering a distance $$S$$ causing similar extension in the spring of constant $$K$$ holding it, $$\mu$$ is the kinetic friction between the block and the surface on which it was moving, the distance $$S$$ is given by:

A
$$\dfrac{1}{K}\mu^2m^2g^2$$

B
$$\dfrac{1}{K}(mKu^2 - \mu^2m^2g^2)^{1/2}$$

C
$$\dfrac{1}{K}(\mu^2m^2g^2 + mK\mu^2 + \mu mg)^{1/2}$$

D
$$\dfrac{\mu mg + \sqrt{\mu^2m^2g^2 + mu^2k}}{k}$$

Solution

By the work-energy theorem $$\dfrac{1}{2}mu^{2}=\mu mgS+\dfrac{1}{2}kS^{2}$$
i.e $$S^{2}+\dfrac{2\mu mgS}{k}-\dfrac{mu^{2}}{k}=0$$
$$\Rightarrow \dfrac{\dfrac{-2\mu mg}{k}+\sqrt{\dfrac{4\mu^{2}m^{2}g^{2}}{k^{2}}+\dfrac{4mu^{2}}{k}}}{2}$$
$$=\dfrac{-\mu mg+\sqrt{\mu^{2}m^{2}g^{2}+mu^{2}k}}{k}$$
Question 9
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Three identical ideal springs, each of spring constant $$K=2000\ N/m$$ are connected in three different arrangements as shown in the figure$$-I$$, figure$$-II$$ and figure$$-III$$ respectively.
A massless hook $$A$$ is connected to the lower end of each configuration. If a mass of $$10\ kg$$ is connected to hook $$A$$, the magnitude of displacement of point $$P$$ are $${x}_{1},x_{2}$$ and $$x_{3}$$ in figure$$-I$$, Figure$$-II$$ and figure$$-III$$ respectively. Choose the correct option.

A
$$x_{1}=0.150\ m,x_{2}=0.075\ m,x_{3}=0.075\ m$$

B
$$x_{1}=0.050\ m,x_{2}=0.025\ m,x_{3}=0.050\ m$$

C
$$x_{1}=0.050\ m,x_{2}=0.025\ m,x_{3}=0.025\ m$$

D
$$x_{1}=0.025\ m,x_{2}=0.050\ m,x_{3}=0.050\ m$$

Solution

$$x_{10}=\dfrac{3mg}{k}$$
$$kx_{1}=\dfrac{k}{2}(x_{10}-x_{1})$$
$$\Rightarrow \quad \dfrac{3kx_{1}}{2}=\dfrac{k}{2}x_{10}$$
$$\Rightarrow \quad 3x_{1}=x_{10}$$
$$\Rightarrow \quad x_{1}=\dfrac{mg}{k}=0.05\ m$$
$$x_{20}=\dfrac{3mg}{2k}$$
$$2kx_{2}=k(x_{20}-x)$$
$$\Rightarrow \quad x_{2}=\dfrac{x_{20}}{3}=\dfrac{mg}{2k}=0.025\ m$$
$$x_{30}=\dfrac{3mg}{2k}$$
$$kx_{3}=2k(x_{20}-x_{3})$$
$$\Rightarrow \quad x_{3}=\dfrac{2x_{20}}{3k}=\dfrac{mg}{k}=0.05\ m$$
Question 10
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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 charged. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along the horizontal axis and momentum is plotted along the `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.
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.

A
$$E_1=\sqrt{2}E_2$$

B
$$E_1=2E_2$$

C
$$E_1=4E_2$$

D
$$E_1=16E_2$$

Solution

From diagram,
Amplitude of oscillator $$1=2\times$$ (Amplitude of oscillator $$2$$)
$$\implies E_1=4E_2$$