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CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 16 - MCQExams.com
CBSE
Class 11 Engineering Physics
Oscillations
Quiz 16
The total spring constant of the system as shown in the figure will be:
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$$\dfrac{k_1}{2} + k_2$$
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$${[\dfrac{1}{2k_1}+\dfrac{1}{k_2}]}^{-1}$$
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$$\dfrac{1}{2k_1} + \dfrac{1}{k_2}$$
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$${[\dfrac{2}{k_1}+\dfrac{1}{k_2}]}^{-1}$$
A particles moves on the x-axis according to the equation $${ x=x }_{ 0 }{ sin }^{ 2 }\omega t$$. The motion is simple harmonic
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With amplitude $${ x }_{ 0 }$$
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With amplitude $${ 2x }_{ 0 }$$
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With time period $$\dfrac { 2\pi }{ \omega } $$
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With time period $$\dfrac { \pi }{ \omega } $$
A particle moves in x-y plane according to the equation $$\overrightarrow {r } =(\hat { i } +2\hat { j })$$ $$A\, cos \,\omega t$$. The motion of the particle is
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simple harmonic
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uniformly accelerated
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circular motion
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projectile motion
A particle executes SHM in accordance with x = A sin $$\omega t.$$ If $${ t }_{ 1 }$$ is the time taken by it to reach from x = 0 to x = $$\sqrt { 3 } (A/2)$$ and $${ t }_{ 2 }$$ is the time taken by it to reach from x = $$\left( \sqrt { 3 } /2 \right) A$$ to x = A the value of $${ t }_{ 1 }/{ t }_{ 2 }$$ is
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2
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$$\dfrac { 1 }{ 2 } $$
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3
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$$\dfrac { 1 }{ 3 } $$
Equation of SHM is $$x=10\sin{10\pi t}$$. Find the distance between the two points where speed is $$50\pi$$ $$cm/sec$$. $$x$$ is in cm and $$t$$ is in seconds
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$$10cm$$
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$$20cm$$
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$$17.32cm$$
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$$5(\sqrt{3}+1) cm$$
The value of phase at maximum displacement from the mean position of a particle in S.H.M. is
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$$\pi /2$$
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$$\pi $$
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Zero
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$$2\pi $$
On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct?
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$$ Y = a sin 2 \pi t/ T - cos 2 \pi t / T $$
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Y = a sin Vt
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$$ y = \dfrac {a}{T} sin \left( \dfrac { Kt }{ a } \right) $$
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$$ Y\quad =\quad a\sqrt { 2 } \left( sin\dfrac { 2\pi t }{ T } -cos\dfrac { 2\pi t }{ T } \right) $$
A block of mass $$1 kg$$ is connected with a massless spring of force constant $$100 N/m$$. At $$t=0$$, a constant force $$F=10 N$$ is applied on the block. The spring is in its natural length at $$t=0$$. The speed of particle at $$x=6cm$$ from mean position is:
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$$4 cm/s$$
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$$10 cm/s$$
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$$80 cm/s$$
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$$50 cm/s$$
The period of oscillation of a simple pendulum of length $$L$$, suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $$\alpha$$, is given by:
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$$2\pi\sqrt{\left(\dfrac{L}{g\cos \alpha}\right)}$$
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$$2\pi\sqrt{\left(\dfrac{L}{g\sin\alpha}\right)}$$
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$$2\pi\sqrt{\left(\dfrac{L}{g}\right)}$$
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$$2\pi\sqrt{\left(\dfrac{L}{g\tan\alpha}\right)}$$
Explanation
$$\textbf{Step 1 - Draw FBD Refer Figure}$$
$$\textbf{Step 2 - Calculation of effective acceleration}$$
$$\bullet$$ Apply Newton second law wrt box
Pseudo force $$F_{P} = mg\sin \theta$$
$$F_{net} = ma$$
$$mg \sin \theta - F_{P} = ma$$
$$mg\sin \theta - mg \sin \theta = ma$$
$$a = 0$$
hence along the incline plane - net acceleration is zero.
But along the normal to the incline plane acceleration is $$a_N=g\cos\theta$$
So net acceleration is $$a_{net} = \sqrt {a^{2} + a_{N}^{2}}$$
$$= \sqrt {0 + (g\cos \theta)^{2}}$$
$$a_{net} = g\cos \theta$$
$$\textbf{Step 3 - Calculation of time period}$$
We know that -
$$T = 2\pi \sqrt {\dfrac {l}{a_{eff}}} = 2\pi \sqrt {\dfrac {l}{g\cos \theta}}$$
$$T = 2\pi \sqrt {\dfrac {l}{g\cos \theta}}$$
Hence option (A) is correct.
The equation of an SHM with amplitude $$A$$ and angular frequency $$\omega$$ in which all the distances are measured from one extreme position and time is taken to be zero at the other extreme position is
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$$x = A \sin \omega t$$
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$$x = A (\cos \omega t + \sin \omega t)$$
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$$x = A - A \cos \omega t$$
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$$x = A + A \cos \omega t$$
Explanation
From given question
$$t = 0$$
$$x = 2A$$
Put $$t = 0, x$$ should be $$2A$$
I) $$x = A\sin 0 = 0 \ X$$
II) $$x = A(\cos 0 + \sin 0) = A \ X$$
III) $$x = A - A\cos 0 = 0 \ X$$
IV) $$X = A + A\cos 0 = A + A = 2A \checkmark$$
A particle executing SHM has max acceleration $$ 50 cm/s^2 $$ , The time period of oscillation of 2 sec and initial displacement 2.5 cm from the equilibrium position.The equation of SHM is $$ (take\ \pi^2 = 10) $$
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$$ X = 2.5 sin \left( \pi t + \frac { \pi }{ 6 } \right) cm $$
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$$ X = 5 sin \left( \frac { \pi t }{ 2 } +\frac { \pi }{ 3 } \right) cm $$
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$$ X = 5 sin \left( \pi t+\frac { \pi }{ 3 } \right) cm$$
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$$X = 5 sin \left( \pi t+\frac { \pi }{ 6 } \right) cm $$
Potential energy of a particle of mass is given by $$U\quad =-\quad { U }_{ 0 }\left( 1-\cos {a x } \right) .$$ which of the following can be a possible expression for its time period
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$$2\pi \sqrt { \frac { m }{ { a }^{ 2 }{ u }_{ 0 } } } $$
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$$2\pi \sqrt { \frac { m }{ { a }{ u }_{ 0 } } } $$
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$$2\pi \sqrt { \frac { ma }{ { U }_{ 0 } } } $$
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$$2\pi \sqrt { \frac { { mu }_{ 0 } }{ a } } $$
Between the plates of the capacitor with potential difference $$V$$ across its plate such that upper plate is $$-ve$$, a ball with positive charge '$$q$$' and mass '$$m$$' is suspended by a thread of length '$$l$$'. If the electrostatic force acting on a ball is less than the gravitational force, what should be the period of the ball?
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$$T=2\pi\sqrt{\left(\dfrac{l}{g}\right)}$$
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$$2\pi\sqrt{\left[\dfrac{1}{\left(g+\dfrac{qE}{m}\right)}\right]}$$
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$$T=2\pi\sqrt{\left[\dfrac{1}{\left(g-\dfrac{qE}{m}\right)}\right]}$$
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$$2\pi\sqrt{\left(\dfrac{lm}{gE}\right)}$$
A particle of mass m moves in a one-dimensional potential energy $$ U(x) = -ax^2 +bx^4 $$, where a and 'b' are positive constants. the angular frequency of small oscillation about the minima of the potential energy is equal to
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$$ \pi \sqrt \frac {a}{2b} $$
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$$ 2 \sqrt \frac {a}{m} $$
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$$ \sqrt \frac {2a}{m} $$
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$$ \sqrt \frac {a}{2m} $$
A particle of mass $$1\ kg$$ is moving where potential energy varies as displacement $$U=10(1-\cos 2x)$$ then:
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time period for $$S.H.M.$$ is $$2\pi\sqrt{\dfrac{1}{20}}$$
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speed of particle is maximum at $$x=0$$
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amplitude of oscillation is $$\dfrac{\pi}{4}$$
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time period for $$S.H.M.$$ is $$2\pi\sqrt{\dfrac{1}{40}}$$
A particle of mass $$2\ kg$$ is moving on a straight line under the action of force $$F=(8-2x)N$$. The particle is released at rest from $$x=6\ m$$. For the subsequent motion select the correct options:
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equilibrium position at $$x=4$$
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amplitude $$=2\ m$$
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time to go from $$x=2\ m$$ to $$4\ m $$ is $$ \pi/2$$
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energy of S.H.M. $$=4\ J$$
Explanation
Now, at equilibrium net force is zero
So, $$F=0$$
$$or, (8-2x) =0$$
$$or, x = 4$$
So, equilibrium position is at $$x=4$$
Now the particle is released from $$x=6$$
So, the amplitude will be $$(6-4)m = 2m$$
Now, $$F = (8-2x)$$
$$or, F = -2(x-4)$$
So, $$k = 2$$
Now, Time period $$T = 2\pi \sqrt{\dfrac{m}{k}}=2\pi \sqrt{2/2} = 2\pi$$
Now, time to go from $$x=2$$ to $$4$$ is $$\dfrac{T}{4}$$ i.e. $$\dfrac{\pi}{2}$$
Now, Energy = $$\dfrac{1}{2}m\omega ^2 A^2 = \dfrac{1}{2}m\dfrac{k}{m} A^2=\dfrac{1}{2}kA^2 = \dfrac{1}{2} \times 2 \times 2^2 = 4 J$$
So, All options are $$correct$$
A block of mass $$m$$ is pushed against a spring whose spring constant is $$k$$ fixed at one end with a wall. The block can slide on a frictionless table as shown in figure. If the natural length of spring is $$L_0$$ and it is compressed to half its length when the block is released, find the velocity of the block, when the spring has natural length.
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$$\sqrt{\dfrac{m}{k_0}}.\dfrac{L_0}{2}$$
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$$\sqrt{\dfrac{3k_0}{m}}.\dfrac{L_0}{2}$$
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$$\sqrt{\dfrac{k_0}{m}}.L$$
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$$\sqrt{\dfrac{k_0L}{m}}$$
A ball is hung vertically by a thread of length $$l$$ from a point $$P$$ of an inclined wall that makes an angle $$\beta$$ with the vertical. The thread with the ball is then deviated through a small angle $$\alpha(\alpha >\beta)$$ and set free. Assuming the wall to be perfectly elastic, the period of such pendulum is:
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$$2\sqrt{\dfrac{l}{g}}\left[\sin^{-1}\left(\dfrac{\beta}{\alpha}\right)\right]$$
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$$2\sqrt{\dfrac{l}{g}}\left[\dfrac{\pi}{2}+\sin^{-1}\left(\dfrac{\beta}{\alpha}\right)\right]$$
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$$2\sqrt{\dfrac{l}{g}}\left[\cos^{-1}\left(\dfrac{\alpha}{\beta}\right)\right]$$
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$$2\sqrt{\dfrac{l}{g}}\left[\cos^{-1}\left(-\dfrac{\beta}{\alpha}\right)\right]$$
Explanation
Deviation of ball with angle $$\alpha$$ generate a motion in ball. we can represent this motion as angular displacement $$\theta=\alpha\sin{(\omega t)}$$
let Time period is $$T$$ without barrier, which is known $$T=2\pi\sqrt{\dfrac{L}{g}}$$
Time taken from mean position to displacement
of angle $$\alpha$$
and back to mean position will be half of $$T$$, $$t_1=\dfrac{T}{2}$$
$$t_1=\dfrac{1}{2}\times 2\pi \sqrt{\dfrac{L}{g}}$$
$$t_1=\pi \sqrt{\dfrac{L}{g}}$$
Let Time taken from mean position to
$$\beta$$
and back to mean position is $$t_2$$
from above formula $$\theta=\alpha\sin{(\omega t)}$$
$$\beta=\alpha\sin{(\omega t_2)}$$
$$\dfrac{t_2}{2}=\dfrac{1}{\omega} \sin^{-1} {\dfrac{\beta}{\alpha}}$$
$$t_2=\dfrac{2}{\omega} \sin^{-1} {\dfrac{\beta}{\alpha}}$$
and $$\omega=\sqrt{\dfrac{L}{g}}$$ so
$$t_2=2\sqrt{\dfrac{L}{g}} \sin^{-1} {\dfrac{\beta}{\alpha}}$$
Total time taken = $$T=t_1+t_2$$
$$T=\pi \sqrt{\dfrac{L}{g}}$$
$$+2\sqrt{\dfrac{L}{g}} \sin^{-1} {\dfrac{\beta}{\alpha}}$$
$$T=2\sqrt{\dfrac{l}{g}}\left[\dfrac{\pi}{2}+\sin^{-1}\left(\dfrac{\beta}{\alpha}\right)\right]$$
$$T=2\sqrt{\dfrac{l}{g}}\left[\cos^{-1}\left(-\dfrac{\beta}{\alpha}\right)\right]$$
Hence option B and D are correct.
The new angular frequency of the system will be:
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$$10\ rad/sec$$
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$$15\ rad/sec$$
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$$20\ rad/sec$$
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$$none\ of\ these$$
Explanation
Angular frequency $$\omega=\sqrt{\dfrac{k}{m}}$$
when new block gets attached with previous one
$$m_f=m_1+m_2$$
$$m_f=13+12=25\,kg$$
$$\omega_f=\sqrt{\dfrac{k}{m_f}}$$
$$\omega_f=\sqrt{\dfrac{2500}{25}}$$
$$\omega_f=\sqrt{100}$$
$$\omega_f=10\,rad/s$$
If an impulse $$J$$ is applied at the centre of oscillation in the plane of oscillation, then angular velocity of the rod will be
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$$\dfrac{4J}{ML}$$
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$$\dfrac{2J}{ML}$$
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$$\dfrac{3J}{2ML}$$
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$$\dfrac{J}{ML}$$
The amplitude of oscillation is
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$$\dfrac{3}{2\pi}m$$
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$$3\ m$$
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$$\dfrac{1}{\pi}m$$
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$$1.5\ m$$$
A spring balance has a scale that can read from $$0$$ to $$50\ kg$$ . The length of the scale is $$20\ cm$$. A body suspended from this balance when displaced and released oscillates harmonically with a time period of $$0.6\ s$$. The mass of the body is ( take $$g=10\ m/s^2)$$
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$$10\ kg$$
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$$25\ kg$$
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$$18\ kg$$
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$$22.8\ kg$$
For the above question locate the centre of oscillation.
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$$\dfrac{L}{4}$$ from $$O$$ (down)
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$$\dfrac{L}{4}$$ from $$O$$ (up)
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$$\dfrac{2L}{3}$$ from $$O$$ (down)
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$$\dfrac{7L}{12}$$ from $$O$$ (down)
New equation for position of the combined body is
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$$(10+3\sin 5t)\ cm$$
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$$(10-3\sin 5t)\ cm$$
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$$(10+3\cos 10t)\ cm$$
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$$(10-3\cos 10t)\ cm$$
Explanation
Equation for position $$x$$ of combined body is given by
$$x=l_{0}+a'\sin (\omega't+\pi)$$
or $$x=\left\{10+3\sin (5t+\pi)\right\}cm$$
or $$x=10-3\sin (5t)cm$$
New amplitude of oscillation is
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$$3\ cm$$
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$$20\ cm$$
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$$10\ cm$$
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$$100\ cm$$
Explanation
Conserving linear momentum.
$$(1+3)v=1\times 0.3+3(-0.3)$$
$$v=-0.15\ m/s$$
Negative sign indicates that combined body starts to move leftward. but at the instant of collision, spring is in its natural length or combined body is in equilibrium position. Hence at $$t=0$$, phase of combined body becomes equal to $$\pi$$.
$$\therefore$$ New amplitude of oscillation is
$$a'=\dfrac{|v|}{\omega}=\dfrac{0.15}{5}=0.03\ m=3\ m$$
A particle performs simple harmonic motion with amplitude $$A$$ and time period $$T$$. The mean velocity of the particle over the time in interval which it travels a distance of $$A/2$$ starting from executing position is
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$$\dfrac{A}{T}$$
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$$\dfrac{2A}{T}$$
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$$\dfrac{3A}{T}$$
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$$\dfrac{A}{2T}$$
Mark out the correct statement(s).
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The block-bullet system performs $$SHM$$ about $$y=mg/k$$
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The block-bullet system performs oscillatory motion but not $$SHM$$ about $$y=mg/k$$.
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The block-bullet system perform $$SHM$$ about $$y=4\ mg/3k$$.
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The block bullet system performs oscillatory motion but not $$SHM$$ about $$y=4\ mg/3k$$
Determine the velocity of particle at $$t=5\ s$$
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$$-0.4\ m/s$$
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$$0.5\ m/s$$
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$$-0.25\ m/s$$
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$$none\ of\ these$$
A particle executing harmonic motion is having velocities $$v_1$$ and $$v_2$$ at distances in $$x_1$$ and $$x_2$$ from the equilibrium position. The amplitude of the motion is
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$$\sqrt{\dfrac{v_1^2x_2-v_2^2x_1}{v_1^2+v_2^2}}$$
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$$\sqrt{\dfrac{v_1^2x_1^2-v_2^2x_2^2}{v_1^2+v_2^2}}$$
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$$\sqrt{\dfrac{v_1^2x_2^2-v_2^2x_1^2}{v_1^2-v_2^2}}$$
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$$\sqrt{\dfrac{v_1^2x_2^2+v_2^2x_1^2}{v_1^2+v_2^2}}$$
The velocity of the particle when it reaches $$B$$ will be
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zero
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$$3\sqrt{gl}$$
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$$2\sqrt{gl}$$
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$$\sqrt{gl}$$
Motion of an oscillating liquid column in a U-tube is
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Periodic but not simple harmonic.
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non-periodic.
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Simple harmonic and time period is independent of the density of the liquid.
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Simple harmonic and time-period is directly proportional to the density of the liquid.
Explanation
(c): Consider a U-tube filled with a liquid of density $$p$$ up to height has shown in the figure. when liquid lifted up to a height y from A to B in arm Q. The liquid level in arm P drops by y from A to C. with the height difference between the two arms are given by 2y.
then $$F=-\rho gA(2y)$$
$$F=ma=(2LA\rho )$$so,
$$a=-\dfrac{g}{L} y$$
so,
The motion of an oscillating liquid column in a U tube is SHM with a period, $$T=2\pi \sqrt{\dfrac{L}{g}}$$,
where L is the height of the liquid column in one arm of U tube is equilibrium position of liquid. Therefore, T is independent of the density of the liquid.
A particle doing simple harmonic motion, amplitude $$= 4\ cm$$, time period $$= 12\ sec$$. The ratio between time taken by it in going from its mean position to $$2\ cm$$ and from $$2\ cm$$ to extreme position is
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$$1$$
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$$1/3$$
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$$1/4$$
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$$1/2$$
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