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CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 10 - MCQExams.com
CBSE
Class 11 Engineering Physics
Oscillations
Quiz 10
A simple pendulum with length $$L$$ and mass $$m$$ of the bob is oscillating with an amplitude $$a$$.
Then the maximum tension in the string is :
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$$mg$$
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$$mg[1+\left(\dfrac{a}{L} \right)^{2}]$$
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$$mg[1+\dfrac{a}{2L}]^{2}$$
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$$mg \left[1+\left(\dfrac{a}{L}\right)\right]^{2}$$
Which of the following is simple harmonic motion ?
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Particle moving in a circle with uniform speed
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Wave moving through a string fixed at both ends
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Earth spinning about its axis
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Ball bouncing between two rigid vertical walls
Explanation
Simple harmonic waves are set up in string sixed at the two ends. So, option $$B$$ is correct.
A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of same dimensions and of low modulus of rigidly $$\eta$$ such that lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force is applied perpendicular to one the side face of A. After the force is withdrawn, block A executes small oscillations, the time period of which is given by
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$$2\pi\sqrt{M\eta L}$$
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$$2\pi\sqrt{\dfrac{M}{\eta L}}$$
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$$2\pi\sqrt{\dfrac{ML}{\eta }}$$
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$$\sqrt{\dfrac{M\eta}{L}}$$
Explanation
Modulus of rigidity, $$\eta=\dfrac{F}{A\theta}$$
Here, $$A=L^2$$ and $$\theta=\dfrac xL$$
Therefore, restoring force is,
$$F=-\eta A\theta=-\eta Lx$$
or
Acceleration, $$a=\dfrac FM=-\dfrac{\eta L}{M} x$$
Since, $$a\propto -x$$, oscillations are simple harmonic in nature, time period of which is given by,
$$T=2\pi \sqrt{|\dfrac{Displacement}{Acceleration}|}=2\pi\sqrt{|\dfrac xa|}$$
$$=2\pi\sqrt{\dfrac{M}{\eta L}}$$
A particle executing SHM of amplitude cm and T=4 s, the time taken by it to move from positive extreme position to half the amplitude is
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1S
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$$\dfrac{1}{3} s$$
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$$\dfrac{2}{3} s$$
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$$\sqrt { \dfrac { 3 }{ 2 } } $$
Explanation
Hence,
option $$(C)$$ is correct answer.
Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is 7: 8, then the ratio of lengths of the two pendulums will be
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7:8
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8:7
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49:64
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64:49
Explanation
Suppose at $$t=0$$, pendulums beings to given ro swing simultaneously.
Hence they will again swing simultaneously
if $$n_1 T_1 =n_2 T_2$$
$$\Rightarrow \dfrac {n_1}{n_2}=\dfrac {T_2}{T_1}=\sqrt {\dfrac {l_2}{l_1}}$$
$$\Rightarrow \dfrac {l_1}{l_2}=\left(\dfrac {n_2}{n_1}\right)^2 =\left(\dfrac {8}{7}\right)^2 =\dfrac {64}{49}$$
A particle is executing shm withh amplitude a and has maximum velocity $$v_0$$ .its speed at displacement $$3a/4$$ will be?
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$$\frac{{{v_0}\sqrt 7 }}{4}$$
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$$\frac{{{v_0}\sqrt 4 }}{4}$$
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$$\frac{{{v_0}\sqrt 7 }}{3}$$
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$$\frac{{{v_0}\sqrt 9 }}{3}$$
Explanation
$$\begin{array}{l} { V_{ \max } }={ V_{ 0 } }=A\omega \\ V=\omega \sqrt { { A^{ 2 } }-{ x^{ 2 } } } \\ =\omega \sqrt { { A^{ 2 } }-{ { \left( { \dfrac { { 3A } }{ 4 } } \right) }^{ 2 } } } \\ =\omega \sqrt { { A^{ 2 } }-\dfrac { { 9{ A^{ 2 } } } }{ { 16 } } } \\ =\omega \sqrt { \dfrac { { 16{ A^{ 2 } }-9{ A^{ 2 } } } }{ { 16 } } } \\ =\omega \sqrt { \dfrac { { 7{ A^{ 2 } } } }{ { 16 } } } \\ =\dfrac { { \omega A } }{ 4 } \sqrt { 7 } \\ =\dfrac { { { v_{ 0 } }\sqrt { 7 } } }{ 4 } \end{array}$$
Hence, Option $$A$$ is correct .
If the frequency of a wave is 20 Hz, its time period will be:
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$$0.05\ s$$
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$$0.5\ s$$
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$$0.2\ s$$
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$$2 s$$
Three masses 700 gm, 500 gm and 400 gm are suspended at the end of the spring and they are in equilibrium. When the 700 gm mass is removed, the system oscillates with a period of 3 sec, when the 500 gm mass is also removed, it will oscillate with a period of
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1 sec
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2 sec
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3 sec
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$$\sqrt { \dfrac { 12 }{ 5 } } sec$$
Explanation
When mass $$700\ gm$$ is removed, the left out mass $$(500+400)\ gm$$ oscillates with a period of $$3\ sec$$
$$\therefore 3=t=2 \pi \sqrt{\dfrac{(500+400)}{k}} ...(i)$$
When $$500\ gm$$ mass is also removed, the left out mass is $$400\ gm$$
$$\therefore t^{'}=2 \pi \sqrt{\dfrac{400}{k}} ...(ii)$$
$$\Rightarrow \dfrac{3}{t^{'}}=\sqrt{\dfrac{900}{400}}$$
$$ \Rightarrow t^{'}=2\ sec$$
A Particle is executing SHM with amplitude A and has a maximum velocity $$v_0$$.
(a) At what displacement will its velocity be $$\cfrac{v_0}{2}$$ ?
(b) What is its velocity at displacement $$A/2$$ ?
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$$
\cfrac { \sqrt { 3 } } { 2 } A , \cfrac{\sqrt{3}} {4} v _ { 0 }
$$
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$$
\cfrac { 1 } { 2 } \mathrm { A } , \cfrac { 1 } { 2 } \mathrm { v } _ { 0 }
$$
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$$
\sqrt { 2 } \mathrm { A } , \cfrac { 3 } { 4 } \mathrm { v } _ { 0 }
$$
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$$
\sqrt { 2 \pi } \mathrm { A } , \cfrac { 3 } { 4 } \mathrm { v } _ { 0 }
$$
A block of mass $$1$$ kg is connected to a spring of spring constant $$\pi^2 N/m$$ fixed at other end and kept on smooth level ground. The block is pulled by a distance of $$1$$ cm from natural length position and released. After what time does the block compress the spring by $$\frac{1}{2} cm$$.
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$$\dfrac{2}{3}$$ sec
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$$\dfrac{1}{3}$$ sec
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$$\dfrac{1}{6}$$ sec
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$$\dfrac{1}{12}$$ sec
wave length of wave is
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$$0.4 m$$
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$$0.2 m$$
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$$0.16 m$$
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$$0.8 m$$
Select the correct statements.
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A simple harmonic motion is necessarily periodic.
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A simple harmonic motion is necessarily oscillatory.
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An oscillatory motion is necessarily periodic.
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A periodic motion is necessarily oscillatory.
Starting from the mean position a body oscillates simple harmonically with a period of $$ 2 s.$$ After what time
will its kinetic energy be $$75 \%$$ of the total energy ?
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$$\dfrac { 1 } { 6 } s$$
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$$\dfrac { 1 } { 4 } s$$
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$$\dfrac { 1 } { 3 } s$$
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$$\dfrac { 1 } { 12 } s$$
Periodic motion of an object or particle is known as
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Vibration
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Amplitude
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frequency
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Time period
A mass m oscillation with simple harmonic motion with frequency $$ f =\frac { \omega }{ 2\pi } $$ and amplitude A on a spring of stiffness constant K. Which of the following is not correct?
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The frequency is $$ \frac {1}{2\pi} \sqrt { \frac { K }{ m } } $$
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The maximum velocity occurs when X = 0
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The total energy of the system is $$ \frac {1}{2} KA^2 $$
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Its minimum potential energy occurs when X = A
A particle is executing the motion $$x = a\cos (\omega t - \theta)$$. The maximum velocity of the particle is
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$$a\omega \cos \theta$$
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$$a\omega$$
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$$a\omega \sin \theta$$
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None of these
A string fixed at both ends, oscillate in $$4^{th}$$ harmonic. The displacement of a particle of string is given as:
$$Y = 2A \sin (5\pi x)\cos (100\pi t)$$. Then find the length of the string?
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$$80cm$$
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$$100cm$$
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$$60cm$$
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$$120cm$$
Explanation
$$\dfrac{2\pi}{\lambda}=5\pi$$
$$4\left(\dfrac{\lambda}{2}\right)= \ell \Rightarrow 2\lambda = \ell \Rightarrow 2\times \dfrac{2}{5} = \ell$$
$$\Rightarrow \ell = 80cm$$
The time period of a simple pendulum is air is $$T$$. Now the pendulum is submerged in a liquid of density $$\dfrac{\rho}{16}$$ where $$\rho$$ is density of the bob of the pendulum. The new time period of oscillation is.
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$$\dfrac{4}{\sqrt{15}} T$$
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$$\sqrt{\dfrac{4}{15}} T$$
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$$\sqrt{\dfrac{15}{4}} T$$
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$$\dfrac{\sqrt{15}}{4} T$$
Explanation
$$T = 2\pi \sqrt{\dfrac{\ell}{g}}$$
$$\tau = I\alpha$$
$$\left(\rho Vg - \dfrac{\rho}{16} Vg\right)\ell \sin \theta = (\rho V)\ell^2 \alpha$$
$$\dfrac{15g}{16\ell} \theta = \alpha$$
$$\omega = \sqrt{\dfrac{15g}{16 \ell}} \Rightarrow T' = 2\pi \sqrt{\dfrac{16\ell}{15g}}$$
$$T' = \dfrac{4}{\sqrt{15}} T$$
The KE and PE of a particle executing S.H.M. will be equal when displacement is:
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a/2
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a/1.414
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1.414a
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1.414/a
A spring of length $$'l'$$ has spring constant $$'k'$$ is cut into two parts of length $$l_{1}$$ and $$l_{2}$$. If their respective spring constants are $$k_{1}$$ and $$k_{2}$$, then $$\dfrac {k_{1}}{k_{2}}$$ is
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$$\dfrac {l_{2}}{l_{1}}$$
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$$\dfrac {2l_{2}}{l_{1}}$$
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$$\dfrac {l_{1}}{l_{2}}$$
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None
Explanation
$$k_{1}l_{1} = k_{2}l_{2} = kl$$
$$\dfrac {k_{1}}{k_{2}} = \dfrac {l_{2}}{l_{1}}$$
The displacement of a damped harmonic oscillator is given by $$x(t)=e^{-01.1t}\, cos (10\pi t+\Phi ).$$ Here $$t$$ is in seconds. The time taken for its amplitude of vibration to drop to half for its initial value is close to :
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$$13s$$
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$$7s$$
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$$27s$$
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$$4s$$
Explanation
$$A=A_0e^{-0.1t}=\dfrac{A_0}{2}$$
$$ln2=0.1t$$
$$t=10ln2=6.93 \approx 7 sec$$
Which of the following equation does not represent a simple harmonic motion:
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$$y=a\sin\omega t$$
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$$y=a\cos\omega t$$
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$$y=a\sin\omega t+b\cos\omega t$$
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$$y= a\tan\omega t$$
Explanation
Standard equation of $$S.H.M$$ $$ \dfrac {d^2 y}{dt^2}=-\omega^2 y$$, is not satisfied by $$y=a\tan \omega t$$
A particle of mass $$10kg$$ is executing S.H.M. of time period $$2$$ second and amplitude $$0.25m$$. The magnitude of maximum force on the particle is:
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$$5N$$
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$$24.65N$$
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Zero
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$$40.6N$$
For a simple pendulum, the graph between $$T^2$$ and $$L$$ is:
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Straight line passing thorugh origin
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Parabolic
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Circle
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None of the above
A spring has a spring constant of $$6.0$$ N $$cm^{-1}$$. It is joined to another spring whose spring constant is $$4.0$$ N $$cm^{-1}$$. A load of $$80$$N is suspended from this composite spring.
What is the extension of this composite spring?
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$$8.0$$cm
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$$16$$cm
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$$17$$cm
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$$33$$cm
Explanation
For series combination: $$x_{eff}=x_1+x_2$$
$$k_1x_1+k_2x_2=F$$
$$x_1=\dfrac{F}{k_1} \ , x_2=\dfrac{F}{k_2}$$
$$x_{eff}=\dfrac{F}{k_1}+\dfrac{F}{k_2}$$
$$x_{eff}=20+13.33=33.33cm$$
$$approx\ x_{eff}=33cm$$
If a particle is acted by two simple harmonic motions simultaneously, the path of particle is:
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Stream line motion
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Lissajous figure
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Just like motion of a particle under gravity
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None of the above
A particle of mass $$m$$ is moving along the X-axis under the potential $$U(x)=\dfrac{kx^2}{2}+{\lambda}{}$$ where $$k$$ and $$\lambda$$ are positive constants of appropriate dimensions. The particle is slightly displaced from its equilibrium position. The particle oscillates with the angular frequency $$(\omega )$$ given by
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$$3\dfrac{k}{m}$$
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$$3\dfrac{m}{k}$$
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$$\sqrt{\dfrac{k}{m}}$$
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$$\sqrt{3\dfrac{m}{k}}$$
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$$\sqrt{3\dfrac{k}{m}}$$
Explanation
Given, $$U(x)=\dfrac{Kx^2}{2}+\lambda$$
We know,
$$F=-\dfrac{\partial u}{\partial x}=\dfrac{-\partial(\dfrac{Kx^2}{2}+\lambda)}{\partial 2}$$
$$=2\dfrac{Kx}{2}+0=-Kx$$
$$a=\dfrac{F}{m}=-\dfrac{K}{m}x$$............(1)
and we also know
$$a=-w^2x$$
compare equation (1) with (ii)
$$w^2=\dfrac{K}{m}$$.........(2)
$$\boxed{w=\sqrt{\dfrac{K}{m}}}$$ Ans ..........(c)
A point mass $$m=20kg$$, is suspended by a massless spring of constant $$2000 N/m$$. The point mass is released when elongation in the spring is $$15cm$$. The equation of displacement of particle as function of time is: (Take $$g=10 m/s^2$$)
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$$y=10\sin 10t$$
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$$y=10\cos 10t$$
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$$y=10\sin\left(10t+\dfrac{\pi}{6}\right)$$
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None of these
Two S.H.M.'s $$x=a\sin\omega t$$ and $$y=b\cos \omega t$$ directed along y-axis respectively are acted on particle. The path of the particle is:
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Circle
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Straight line
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Ellipse
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Parabola
A particle moves along y-axis according to equation $$y=3+4\cos \omega t$$. The motion of particle is:
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Not S.H.M.
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Oscillatory but not S.H.M.
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S.H.M.
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None of the above
A naughty boy is sitting on the roof of a flat toy car of mass $$6$$ kg. If no slipping takes place between car and the boy then what should be the mass of the child in order to have period of system equal to $$0.758$$ sec?
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$$2.74$$ kg
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$$6$$ kg
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$$3$$ kg
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None of these
A spring of spring constant $$200N/m$$ has a block of mass $$1 kg$$ hanging at its one end and of other end spring is attached to a ceiling of an elevator. The elevator rising upwards with an acceleration of $$g/3$$. When acceleration is suddenly cease, then what should be the angular frequency and elongation during the time when the elevator is accelerating?
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$$14.14\ rad/s, \, 0.07m$$
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$$13\ rad/s, \, 0.1m$$
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$$14\ rad/s, \, 0.05m$$
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$$10\ rad/s, \, 0.07m$$
There are two pendulums of length $$l_1$$ and $$l_2$$ which start vibrating. At some instant, the both pendulum are in mean position in the same phase. After how many vibrations of shorter pendulum, the both pendulum will be in phase in the mean position?
$$[(l_1 > l_2), l_1=121 cm, l_2=100 cm]$$
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$$11$$
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$$10$$
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$$9$$
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$$8$$
A body with speed '$$v$$' is moving along a straight line. At the same time it is at distance $$x$$ from a fixed point on the line, the speed is given by $$v^2=144-9x^2$$. Then:
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Displacement of the body $$<$$ distance moved by body
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The magnitude of acceleration at a distance $$3$$m from the fixed point is $$27 m/s^2$$
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The motion is S.H.M. with $$T=\dfrac{2\pi}{3}$$ unit
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The maximum displacement from the fixed point is $$4$$ unit
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All the above
The equations of two linear S.H.M.'s are
$$x=a\sin \omega t$$ along x-axis
$$y=a\sin 2\omega t$$ along y-axis
If they act on a particle simultaneously, the trajectory of the particle is:
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$$\dfrac{y^2}{a^2}+\dfrac{x^2}{4a^2}=1$$
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$$y^2=\dfrac{4x^2}{a^2}(a^2-x^2)$$
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$$y^2=2ax$$
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None of these
The motion of a particle is given by
$$y=4\sin\omega t+8\sin \left(\omega t+\dfrac{\pi}{3}\right)$$
The motion of particle is:
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S.H.M.
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Not S.H.M.
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Periodic but not S.H.M.
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None of the above
There is a clock which gives correct time at $$20^o$$C is subjected to $$40^o$$C. The coefficient of linear expansion of the pendulum is $$12\times 10^{-6}$$ per $$^oC$$, how much is gain or loss in time?
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$$10.3$$ sec/day
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$$19$$ sec/day
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$$5.5$$ sec/day
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$$6.8$$ sec/day
A particle of mass $$m$$ moves in a straight line. If $$v$$ is the velocity at a distance $$x$$ from a fixed point on the line and $$v^{2}=a-bx^{2}$$, where $$a$$ and $$b$$ are constant then:
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the motion continues along the positive $$x-$$ direction only
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the total energy of the particle is $$$ma$$
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the particle oscillates with a frequency equal to $$\dfrac{\sqrt{b}}{2\pi}$$
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the motion is simple harmonic
Explanation
Given,
$$v^2 = a-bx^2$$
$$\therefore\, v^2 + bx^2 = a$$
$$\dfrac {x^2}{a/b} + \dfrac {v^2}{a} = 1$$
For a S.H.M:
$$x= Asin(\omega t+ \theta)$$
$$v= A\omega cos(\omega t+ \theta)$$
$$\dfrac {x^2}{A^2} + \dfrac {v^2}{A^2\omega^2} = 1$$
Comparing both equations:
$$a/b= A^2$$
$$a= A^2\omega^2$$
$$\therefore\, b= \omega^2$$
$$\Rightarrow\, \omega=\sqrt{b}$$
$$\Rightarrow\, f= \dfrac {\omega}{2\pi} = \dfrac {\sqrt{b}}{2\pi}$$
And, the motion is simple harmonic
A spring-mass system oscillation in a car. if the car accelerates on a horizontal road, the frequency of oscillation will
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increase
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decrease
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remain same
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become zero
Explanation
The frequency of a spring-mass system depends on the mass attached to the spring and the spring constant $$K$$ by the equation:
$$\omega = \sqrt{\dfrac Km}$$
Angular frequency is independent of acceleration of system ,Hence remains same.
Option C is correct.
A particle of mass m is allowed to oscillate near the minimum of a vertical parabolic path having the equation $$x^2=4ay$$. The angular frequency of small oscillations is given by:
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$$\sqrt{gh}$$
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$$\sqrt{2gh}$$
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$$\sqrt{\left(\dfrac{g}{2a}\right)}$$
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$$\sqrt{\left(\dfrac{g}{a}\right)}$$
Figure represents two simple harmonic motions.
The parameter which has different value in the two motions is
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amplitude
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frequency
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phase
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maximum velocity
An elastic ball of density $$d$$ is released and it falls through a height $$h$$ before striking the surface of liquid of density $$\rho(d < \rho)$$. The motion of ball is:
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Periodic
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S.H.M.
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Circular
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Parabolic
Explanation
The density of the ball is less than the density of the liquid. So it will float in the liquid and a buoyant force will act on the body which will decelerate the body and after going some deep inside the liquid the buoyant force will be greater than the gravitational force and the body will start to go up. In this way it will execute periodic motion.
The displacement of a particle in simple harmonic motion in one time period is
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$$A$$
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$$2A$$
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$$4A$$
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zero
Explanation
During a Simple harmonic motion, in one oscillation the final and initial point of particle is at same position . Thus there is no displacement of particle .
Option D is correct.
$$m_1$$ and $$m_2$$ are connected with a light inextensible string with $$m_1$$ lying in smooth table and $$m_2$$ hanging as shown in figure. $$m_1$$ is also connected to a light spring which is initially unstretched and the system is released form rest:
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system performs S.H.M with angular frequency given by $$\sqrt{\dfrac{k(m_1+m_2)}{m_1m_2}}$$
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maximum displacement of $$m_1$$ will be $$\dfrac{m_2g}{k}$$
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tension in string will be $$0$$ when the system is released
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system performs S.H.M with angular frequency given by $$\sqrt{\dfrac{k}{m_1+m_2}}$$
A hollow sphere is filled with water. It is hung by a long thread to make it a simple pendulum. As the water flows out of a hole at the bottom of the sphere, the frequency of oscillation will
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go on increasing
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go on decreasing
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first increases and then decreases
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first decreases and then increases
Explanation
Frequency of a simple pendulum
$$f=\dfrac {1}{2\pi}\sqrt {\dfrac gl}$$
Evidently, $$f\infty \dfrac {1}{\sqrt l}$$ at any place.
The effective length $$l$$ is the distance of the point of suspension $$O$$' from the cenre of gravity $$(CG)$$ of the bob. As, water flows out of the hole at the bottom, the $$CG$$ descends from centre towards the bottom, increasing the effective length and consequently $$f$$ decreases.]
However, when all the water has flows out the $$CG$$ of a hollow sphere, is once again at its centre and hence the effective.
The displacement of two identical particles executing S.H.M are represented by equations
$$x_1=4\sin \left( 10t+\dfrac{\pi}{6}\right)$$ and $$x_2=5\cos \omega t$$
For what value of $$\omega$$ energy of both the particles is same?
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$$\dfrac{2}{\sqrt 2}$$ unit
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$$4$$ unit
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$$2\sqrt 2$$ unit
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$$8$$ unit
Explanation
Energy by a particle in SHM is given by
$$E=\dfrac{1}{2}m\omega^2A^2$$; where A is amplitude.
Given amplitude $$A_1=4$$ and
$$A_2=5$$
$$\omega_1=10$$ and
$$\omega_2=\omega$$
$$E_1=\dfrac{1}{2}m_1\omega_1^2A_1^2$$
$$E_2=\dfrac{1}{2}m_2\omega_2^2A_2^2$$
according to question
$$E_1=E_2$$
$$\dfrac{1}{2}m_1\omega_1^2A_1^2$$
$$=\dfrac{1}{2}m_2\omega_2^2A_2^2$$
putting values
$$\dfrac{1}{2}m10^2\times4^2$$
$$=\dfrac{1}{2}m\omega_2^25^2$$
$$1600=25\omega^2$$
$$\omega=8$$ unit
A particle of mass $$m$$ moving along the $$x-$$ axis as a potential energy $$U(x)=a+bx^2$$ where $$a$$ and $$b$$ are positive constants. It will execute simple harmonic motion with a frequency determined by the value of :
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$$b$$ alone
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$$b$$ and $$a$$ alone
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$$b$$ and $$m$$ alone
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$$b,a $$ and $$m$$ alone
Explanation
$$U=a+bx^2$$
$$F=-\dfrac {dU}{dx}=-[0+2bx]$$
$$F=ma=-2bx \ \Rightarrow \ a=-\left(\dfrac {2b}{m}\right)x$$
$$\omega =2\pi f =\left(\dfrac {2b}{m}\right)^{1/2}$$
Hence frequency depends upon $$b$$ and $$m$$
A block of mass $$m$$ is suspended by a rubber cord of natural length $$l=mg/k$$, where $$k$$ is force constant of the cord. The block is lifted upwards so that the cord becomes just tight and then block is released suddenly. Which of the following will not be true?
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Block performs periodic motion with amplitude greater than $$l$$
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Block performs $$SHM$$ with amplitude equal to $$l$$
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Block will never return to the position from where it was released
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Angular frequency $$\omega$$ is equal to $$1\ rad/s$$.
Explanation
When the block is released suddenly, it starts to move down. During its downward motion the rubber cord elongates. Hence, a tension is developed in it but the block continues to accelerate downwards till tension becomes equal to weight $$mg$$ of the block.
After this moment, the block continues to move down due to its velocity and rubber cord further elongates. Therefore, tension becomes greater than the weight; hence the block now retards and comes to an instantaneous rest.
At lowest position of the block, strain energy in the cord equals loss of potential energy of the block. Suppose the block comes to an instantaneous rest when elongation of the rubber cord is equal to $$y$$. Then
$$\dfrac{1}{2}ky^2=mgy\Rightarrow y=\dfrac{2mg}{k}$$ and $$0$$,
Hence, block will be instantaneously at rest at $$y=0$$ and at $$y=2mg/k$$.
In fact, the block oscillates between these two values.
Since the rubber cord is elastic, tension in it is directly proportional to elongation. Therefore, the block will perform SHM.
Its amplitude will be equal to half of the distance between these extreme positions of the blocks or amplitude
$$=\dfrac{1}{2}\times \dfrac{2mg}{k}=\dfrac{mg}{k}=l$$
Hence, option (b) is correct.
The angular frequency of its SHM will be equal to
$$\omega=\sqrt{\dfrac{k}{m}}$$
Since $$k$$ and $$m$$ are not given in the question, it cannot be calculated. Hence, option (d) is not correct.
A simple pendulum consists of a bob of mass $$m$$ and a light string of length $$l$$ as shown in the Fig $$4.138$$, Another identical ball moving with the small velocity $$v_{0}$$ collides with the pendulum bob and sticks to it. For this new pendulum of mass $$2m$$ mark out the correct statement(s)
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Time period of the pendulum is $$2\pi\sqrt{\dfrac{l}{g}}$$
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The equation of motion for this pendulum is $$\theta=\dfrac{v_{0}}{2\sqrt{gl}}\sin \left[\sqrt{\dfrac{g}{l}}t\right]$$
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The equation of motion for this pendulum is $$\theta=\dfrac{v_{0}}{2\sqrt{gl}}\cos \left[\sqrt{\dfrac{g}{l}}t\right]$$
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Time period of the pendulum is $$2\pi\sqrt{\dfrac{2l}{g}}$$
Explanation
The time period of the simple harmonic pendulum is independent of mass, so it would be same as that $$T=2\pi \sqrt{l/g}$$.
After a collision, the combined mass acquires a velocity of $$v_0/2$$, as a result of this velocity, the mass $$(2m)$$ moves up and at angel $$\theta_0$$ (say) with vertical, it stops, this is the extreme position of bob.
From work-energy theorem, $$\Delta K=W_{total}$$
$$0-\dfrac{2m}{2}\left( \dfrac{v_0}{2}\right)^2=-2mgl(1-\cos \theta_0)$$
$$\dfrac{v_0^2}{8gl}=1-\cos \theta =2\sin^2\dfrac{\theta_0}{2}$$
$$\sin \dfrac{\theta_0}{2}=\dfrac{v_0}{4\sqrt{gl}}$$
If $$\theta_0$$ is small, $$\sin \dfrac{\theta_0}{2}=\dfrac{\theta_0}{2}\Rightarrow \theta_0=\dfrac{v_0}{2\sqrt{g\ell}}$$
So, the equation of simple harmonic motion is $$\theta=\theta_0\sin ( \omega t)$$
A simple pendulum has a time period $$T$$. The bob is now given positive charge:
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if some positive charge is placed at the point of suspension $$T$$ will increase
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if some positive charge is placed at the point of suspension, $$T$$ will not change
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if a uniform downward electric field is switched on $$T$$ will increase
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if a uniform downward electric field is switched on $$T$$ will decrease
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