Oscillations Test

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Oscillations Test - 62

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Question 1
1 / -0

An elastic string of length $$\displaystyle l $$ supports a heavy particle of mass m and the system is in equilibrium with elongation produced being e as shown in the figure. The particle is now pulled down below the equilibrium position through a distance $$\displaystyle d\left ( \leq e \right )$$ and released. The angular frequency and maximum amplitude for SHM is

A
$$\displaystyle \sqrt{\frac{g}{e}},e$$

B
$$\displaystyle \sqrt{\frac{g}{l }},2e$$

C
$$\displaystyle \sqrt{\frac{g}{d+e}},d$$

D
$$\displaystyle \sqrt{\frac{g}{e}},2d$$

Solution

Here, weight of the ball must balance spring force at equilibrium position.
$$kx=mg\\ ke=mg\\ k=\frac { mg }{ e } $$
After elongating more, the net restoring force will be,
$$F=kd\\ a=\frac { Kd }{ m } \\ a={ \omega }^{ 2 }x\\ \omega =\sqrt { \frac { k }{ m } } \\ =\sqrt { \frac { mg }{ em } } \\ =\sqrt { \frac { g }{ e } } \\ T=\frac { 2\pi }{ \omega } =2\pi \sqrt { \frac { m }{ k } } $$
Now, putting the value of K
$$T=2\pi \sqrt { \frac { m\times e }{ mg } } \\ T=2\pi \sqrt { \frac { e }{ g } } \\ g={ \omega }^{ 2 }A$$ [At extreme position]
$$g=\frac { g }{ e } \times A\\ A=e$$
Question 2
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A practice is executing SHM. its time period is equal to the smallest time interval in which the particle acquires a particular velocity,$$\bar { v } $$, the magnitude of $$\bar { v } $$ may be:

A
Zero

B
$$\displaystyle V_{max}$$

C
$$\displaystyle \frac{V_{max}}{2}$$

D
$$\displaystyle \frac{V_{max}}{\sqrt{2}}$$
Question 3
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A particle is performing SHM of amplitude "A" and time period "T". Find the time taken by the particle to go from 0 to A /$$\sqrt{2}$$.

A
T/12

B
T/$$\sqrt{2}$$

C
T/6

D
t/4

Solution

Let equation of SHM be x = A $$\displaystyle \sin \omega t$$
when x = 0 t = 0 when x = A/2 ; A/2 = A $$\displaystyle \sin \omega t$$
or $$\displaystyle \sin \omega t=1/2$$ $$\displaystyle \omega t=\pi /6$$
$$\displaystyle \frac{2\pi }{T}t=\pi /6$$ t=T/12
Hence time taken is T/12 where T is time period of SHM
Question 4
1 / -0

Tow identical springs are connected to mass $$m$$ as shown ($$k=$$ spring constant). IF the period of the configuration in (a) is $$2s$$, the period of the configuration in (b) is

A
$$\sqrt {2}s$$

B
$$1s$$

C
$$\cfrac{1}{\sqrt {2}}s$$

D
$$2\sqrt {2}s$$

Solution

Series combination :
Equivalent spring constant:
$$\Rightarrow k_s = \dfrac{k_1 \times k_2}{k_1+k_2} = \dfrac{k \times k}{k+k} = \dfrac{k}{2}$$
Time period, $$T_s =2\pi \sqrt{\dfrac{m}{k_s}} = 2\pi \sqrt{\dfrac{2m}{k}}$$ ............(1)

Parallel combination :
Equivalent spring constant:
$$\Rightarrow k_p = k_1 +k_2 = k+k = 2k$$
Time period, $$T_p =2\pi \sqrt{\dfrac{m}{k_p}} = 2\pi \sqrt{\dfrac{m}{2k}}$$ ............(2)
$$\Rightarrow $$$$\dfrac{T_p}{T_s} = \dfrac{1/\sqrt{2}}{\sqrt{2}} = \dfrac{1}{2}$$
$$\Rightarrow \dfrac{T_p}{2} = \dfrac{1}{2}$$
$$\Rightarrow T_p = 1\ s$$
Question 5
1 / -0

Two identical springs of spring constant $$k$$ are connected in series and paralIeI as shown in figure. $$A$$ mass $$M$$ is suspended from them. The ratio of their frequencies of vertical oscillation will be :

A
$$1 : 2$$

B
$$2 : 1$$

C
$$4 : 1$$

D
$$1 : 4$$

Solution

We know that: $$\displaystyle T=2\pi \sqrt {\frac {m}{K}} \Rightarrow n= \frac {1}{2\pi} \sqrt {\frac {K}{m}}$$
For a spring mass system.
In case 1: If $$K$$ is the resultant spring constant, then$$\displaystyle \frac {1}{K} = \frac {1}{k} + \frac {1}{k} = \frac {2}{k} \Rightarrow K = \frac {k}{2}$$
In case 2, $$K = k + k =2k$$
If $$n_1$$ & $$n_2$$ be frequencies in two cases, then
$$\displaystyle n_1 = \frac {1}{2\pi} \sqrt {\frac {k}{2m}}; n_2 = \frac {1}{2\pi} \sqrt {\frac {2k}{m}};$$
$$\displaystyle \Rightarrow \frac {n_1}{n_2} = \sqrt {\frac {1}{4}} \Rightarrow \frac {n_1}{n_2} = \frac {1}{2}$$
Question 6
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A particle executing simple harmonic motion has an angular frequency of 6.28 $$\displaystyle s^{-1}$$ and amplitude of 10 cm, find the speed at t = 1/6 s assuming that the motion starts from rest at t = 0.

A
$$54.4\: cm/s$$

B
$$44.4\: cm/s$$

C
$$64.4\: cm/s$$

D
$$14.4\: cm/s$$

Solution

At t = 0 the velocity is zero i.e. the particle is at an extreme The equation for displacement may be written as $$\displaystyle x=A\cos \omega t$$
The velocity is $$\displaystyle v=-A\omega \sin \omega t$$At $$\displaystyle t=\frac{1}{6}s$$ $$\displaystyle v=-(0.1m)(6.28s^{-1})\sin \left ( \frac{6.28}{6} \right )$$ $$\displaystyle =(-0.628\: m/s)\sin \frac{\pi }{3}=54.4\: cm/s$$
Question 7
1 / -0

A particle is moving on a circular track with uniform speed. Its motion is

A
Periodic and simple harmonic

B
Periodic but not simple harmonic

C
Damped

D
None of these.
Question 8
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The left block in figure collides inelastically with the right block and sticks to it. The amplitude of the resulting simple harmonic motion is

A
$$\sqrt{\dfrac{m}{2k}}v$$

B
$$\sqrt{\dfrac{m}{k}}v$$

C
$$\sqrt{\dfrac{2m}{k}}v$$

D
$$\sqrt{\dfrac{m}{4k}}v$$

Solution

Assuming the collision to last for a small interval only we can apply the principle of conservation of momentum The common velocity after the collision is $$\displaystyle \frac{v}{2}$$ The kinetic energy = $$\displaystyle \frac{1}{2}(2m)\left ( \frac{v}{2} \right )^{2}=\frac{1}{4}mv^{2}$$ This is also the total energy of vibration as the spring is unstretched at this moment If the amplitude is A the total energy can also be written as $$\displaystyle \frac{1}{2}kA^{2}$$ Thus $$\displaystyle \frac{1}{2}kA^{2}=\frac{1}{4}mv^{2}$$ giving $$\displaystyle A=\sqrt{\frac{m}{2k}}v$$
Question 9
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A simple harmonic oscillator consist of a block attached to a spring with k = 100 N/m. The block slides on a frictionless horizontal surface, with equilibrium point x = 0. A graph of block's velocity as a function of time is shown. correctly match the two column:- ($$\pi^2 = 10)$$
List - I List - II
(P) The block's mass in Kg (1) -0.20
(Q) The block's displacement at t = 0 in meters (2) -200
(R) The block's acceleration at t = 0.1 s in $$m/s^2$$ (3) 0.20
(S) The block's maximum kinetic energy in joules (4) 4.0

A
P-1 Q-3 R-2 S-4

B
P-4 Q-2 R-3 S-1

C
P-3 Q-4 R-1 S-2

D
P-2 Q-1 R-4 S-3

Solution

$$(P)\quad T=2\pi \sqrt { \dfrac { m }{ K } } \\ 0.1=2\times 3.14\sqrt { \dfrac { m }{ 100 } } \\ m={ \left( \dfrac { 0.1 }{ 2\times 3.14 } \right) }^{ 2 }\times 100\\ m=0.025Kg$$

$$(Q)$$ From above graph we can get,
$$v=\pi sin\left( \sqrt { \dfrac { K }{ m } } t \right) \\ v=\pi sin\left( \sqrt { \dfrac { 100 }{ 0.025 } } t \right) \\ v=\pi sin\left( 20\sqrt { 10 } t \right) \\ \dfrac { dx }{ dt } =\pi sin\left( 20\sqrt { 10 } t \right) \\ \int _{ o }^{ x }{ dx= } \int _{ 0 }^{ t }{ \pi } sin\left( 20\sqrt { 10 } t \right) dt\\ x=\dfrac { \pi }{ 20\sqrt { 10 } } cos\left( 20\sqrt { 10 } t \right) $$
at $$t=0 x=-\dfrac { \pi }{ 20\sqrt { 10 } } =0.05m$$

$$(R)\quad a=\dfrac { dv }{ dt } =\pi \sqrt { 10 } \times 20cos\left( 20\sqrt { 10 } t \right) \\ at\quad t=0.1\\ a=198.6\approx 200m/{ s }^{ 2 }$$

$$(S)\quad { V }_{ max }=\pi \quad [when\quad sin\theta =1]\\ { KE }_{ max }=\dfrac { 1 }{ 2 } \times 0.025\times { \pi }^{ 2 }\\ =0.125Joule$$
Question 10
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A particle is oscillating simple harmonically with angular frequency $$\omega$$ and amplitude $$A$$. It is at a point (A) at certain instant (shown in figure). At this instant it is moving towards mean positive (B). It takes time $$t$$ to reach position (B). If time period of oscillations is $$T$$, the average speed between $$A$$ and $$B$$ is :

A
$$\dfrac {A\sin \omega t}{t}$$

B
$$\dfrac {A\cos \omega t}{t}$$

C
$$\dfrac {A\sin \omega t}{T}$$

D
$$\dfrac {A\cos \omega t}{T}$$

Solution

$$x = A\sin \omega t$$

$$dx = A\omega \cos \omega t \ dt$$

$$(v) = \dfrac {\int dx}{\int dt}$$

$$= \dfrac {\int_{0}^{t} A\omega \cos \omega t dt}{\int_{0}^{t} dt}$$

$$= \dfrac {A\omega \dfrac {\sin \omega t}{\omega}}{t}$$

$$= \dfrac {A\sin \omega t}{t}$$.