Oscillations Test

Result

Oscillations Test - 13

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The displacement x of a particle in motion is given in terms of time by x (x - 4) = 1 - 5 $$\cos \omega t$$

A
The particle executes SHM

B
The particle executes oscillatory motion which is not SHM

C
motion of the particle is neither oscillator; nor simple harmonic

D
The particle is not acted upon by a force when it is at x = 4

Solution

$$\begin{array}{l} x\left( { x-u } \right) =1-5\cos wt \\ { x^{ 2 } }-4x=1-5\cos wt \\ x=\dfrac { { 4\pm \sqrt { 16+4\left( { 1-5\cos wt } \right) } } }{ 2 } \\ =\dfrac { { 4\pm \sqrt { 20-2\cos wt } } }{ 2 } \\ =\dfrac { { 4\pm \sqrt { 20\left( { 2{ { \sin }^{ 2 } } } \right) wt } } }{ 2 } \\ =\dfrac { { 4\pm \sqrt { 40 } \sin wt } }{ 2 } \\ =2\pm \sqrt { 10 } \sin wt \\ So.\, particle\, perform\, SHM \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$$
Question 2
1 / -0

Two particle execute S H M of same amplitude and frequency along the same straight line, They pass one another going in opposite direction each time their displacement is half of their amplitude. The phase difference between them is

A
$$0 ^ { \circ }$$

B
$$120 ^ { \circ }$$

C
$$180 ^ { \circ }$$

D
$$135 ^ { \circ }$$

Solution

$$\begin{array}{l} According\, to\, \, question............. \\ In\, first\, case: \\ \, \, \, \, \, \, \frac { a }{ 2 } =a\, sin\, \omega t \\ \Rightarrow sin\, \omega t=\frac { 1 }{ 2 } \\ and\, in\, \, { { second } }\, \, case: \\ \, \, \, \, \, \, \frac { a }{ 2 } =a\, sin\, (\omega t+\varphi ) \\ \Rightarrow sin\, (\omega t+\varphi )=\frac { 1 }{ 2 } \\ Now, \\ \Rightarrow sin\, \omega t\cos \varphi +\cos \omega t\sin \phi =\frac { 1 }{ 2 } \\ \Rightarrow \frac { 1 }{ 2 } \cos \varphi +\sqrt { 1-\frac { 1 }{ 4 } } \sin \varphi =\frac { 1 }{ 2 } \\ \Rightarrow \sqrt { 3 } \sin \phi =1-\cos \varphi \\ \Rightarrow { \sin ^{ 2 } }\varphi =1+{ \cos ^{ 2 } }\varphi -2\cos \varphi \\ \Rightarrow 4{ \cos ^{ 2 } }\varphi -2\cos \varphi -2=0 \\ \Rightarrow \cos \varphi =\frac { { 1-3 } }{ 4 } =-\frac { 1 }{ 2 } \\ \therefore \, \, \, \, \, \cos \varphi =-\frac { 1 }{ 2 } \\ \, \, \, \, \, \, \, \, \, \, \varphi ={ 120^{ 0 } } \\ so\, \, that\, \, the\, correct\, \, option\, is\, B. \end{array}$$
Question 3
1 / -0

A uniform solid sphere of mass m and radius R is suspended in vertical plane from a point on its periphery.The time period of its oscillation is:

A
$$2 \pi \sqrt {\cfrac R g}$$

B
$$2 \pi \sqrt {\cfrac {3R} {2g}}$$

C
$$2 \pi \sqrt {\cfrac {7R} {5g}}$$

D
$$2 \pi \sqrt {\cfrac {2R} {5g}}$$

Solution

$$T=2\pi \sqrt { \cfrac { I }{ mgL } } $$
where $$I=m{ R }^{ 2 }+\cfrac { m{ R }^{ 2 } }{ 2 } =\cfrac { 3m{ R }^{ 2 } }{ 2 } $$
$$\therefore T=2\pi \sqrt { \cfrac { 3m{ R }^{ 2 } }{ 2mgR } } =2\pi \sqrt { \cfrac { 3{ R } }{ 2g } } $$
Question 4
1 / -0

Function $$x = A \sin ^ { 2 } \omega t + B \cos ^ { 2 } \omega t + C \sin \omega t \cos \omega t$$ represents SHM

A
For any value of A ,B and C (except C=0)

B
$$A = - B , C = 2 B , \text { amplitude } = | B \sqrt { 2 } |$$

C
$$A = B ; C = 0$$

D
$$A = B ; C = 2 B , \text { amplitude } = | B |$$

Solution

$$\begin{array}{l} x=A{ \sin ^{ 2 } }\omega t+B{ \cos ^{ 2 } }\omega t+C\sin \omega \cos \omega t \\ =\dfrac { { A\left( { 1-\cos 2\omega t } \right) } }{ 2 } +\dfrac { { B\left( { \cos 2\omega t-1 } \right) } }{ 2 } +\dfrac { C }{ 2 } \sin 2\omega t \\ =\left( { \dfrac { A }{ 2 } -\dfrac { B }{ 2 } } \right) +\left( { \dfrac { B }{ 2 } -\dfrac { A }{ 2 } } \right) \cos \omega t+\dfrac { c }{ 2 } \sin \omega t \\ if\, \, \, \, A=B\, \, \, \, \, \, \, \& \, \, \, C=2B \\ x=B+B\sin 2\omega t \\ \therefore Amplitude\, \, =B \end{array}$$
$$\therefore $$ Option $$D$$ is correct.
Question 5
1 / -0

For a simple pendulum, graph between $$L$$ and $$T$$ will be a :

A
hyperbola

B
parabola

C
straight line

D
a curve line

Solution

$$T=2\pi \sqrt{\dfrac{l}{g}}$$
$$\Rightarrow l\propto T^{2}$$ (Equation of parabola)
Question 6
1 / -0

The phase difference between the displacement and acceleration of particle executing S.H.M. in radian

A
$$\pi/4$$

B
$$\pi/2$$

C
$$\pi$$

D
$$2\pi$$

Solution

SHM is represented by $$x=A\sin\omega t$$
Then, the velocity and acceleration after differentiation will be
$$v=A\omega \cos\omega t$$
$$a=-A\omega^2\sin\omega t$$
Now, we see displacement is in $$\sin\omega t$$ & velocity is in $$\cos\omega t$$
$$\Rightarrow$$ Phase difference between x & $$v=\pi/2$$ rad
Similarly, we see velocity in $$\cos\omega t$$ & acceleration in $$\sin \omega t$$
$$\Rightarrow$$ Phase difference between v & $$a=\pi/2$$rad
Thus, we can conclude that the phase difference between x & $$a=\pi$$ rad.
Question 7
1 / -0

A particle of mass $$2 kg$$ moves along x - axis with potential energy depending upon 'x' co-ordinate as $$ U_(x) = (x^2 - 2 x ) J$$. The mass is $$2 kg$$ .

A
SHM with period $$\pi$$ second

B
SHM with period $$2 \pi$$ second

C
oscillatory but not SHM

D
SHM with period $$\frac{1}{2\pi}$$ second
Question 8
1 / -0

The natural angular frequency of a particle of mass 'm' attached to an ideal spring of force constant 'K' is

A
$$\sqrt{\frac{K}{m}}$$

B
$$\sqrt{\frac{m}{K}}$$

C
$$\left ( \frac{K}{m} \right )^{2}$$

D
$$\left ( \frac{m}{K} \right )^{2}$$

Solution

Suppose you displace the particle by a distance $$'x'$$
The spring now exerts a force,
This provides nccenary force for $$SHM$$
$$\Rightarrow \ F=mwe^2x=k2$$ ($$w:$$ natural angular frequency )
$$\Rightarrow \ w=\sqrt {K/m}$$
Question 9
1 / -0

The motion of a pendulum is

A
rotatory

B
oscillatory

C
curvilinear

D
rectilinear

Solution

The motion of a pendulum is oscillatory.
Question 10
1 / -0

The equation of motion of a particle is $$x = a \cos (a t)^{2}$$. The motion is

A
periodic but not oscillatory.

B
Periodic and oscillatory.

C
oscillatory but not periodic

D
neither periodic nor oscillatory

Solution

$$x = a \cos (a t)^{2}$$ here,

$$x$$ is a cosine.

This results in oscillatory motion.

$$f(t)= f(x)+ f(t)= f(x+t)$$

$$x(t)= a \cos \alpha {t} ^{2}$$

$$x(t+T)= a \cos( \alpha {t+T})^{2}$$

$$ a\cos(\ alpha {t}^{2} )+ (\alpha {T}^{2}) +(2\alpha t T)$$

which is not equal to $$x(t)$$ this shows this is not periodic.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Oscillations Test - 13

Result

Oscillations Test - 13

Score

-

Rank

-

SHARING IS CARING

Question 1

The particle executes SHM

The particle executes oscillatory motion which is not SHM

motion of the particle is neither oscillator; nor simple harmonic

The particle is not acted upon by a force when it is at x = 4

Solution

Question 2

$$0 ^ { \circ }$$

$$120 ^ { \circ }$$

$$180 ^ { \circ }$$

$$135 ^ { \circ }$$

Solution

Question 3

$$2 \pi \sqrt {\cfrac R g}$$

$$2 \pi \sqrt {\cfrac {3R} {2g}}$$

$$2 \pi \sqrt {\cfrac {7R} {5g}}$$

$$2 \pi \sqrt {\cfrac {2R} {5g}}$$

Solution

Question 4

For any value of A ,B and C (except C=0)

$$A = - B , C = 2 B , \text { amplitude } = | B \sqrt { 2 } |$$

$$A = B ; C = 0$$

$$A = B ; C = 2 B , \text { amplitude } = | B |$$

Solution

Question 5

hyperbola

parabola

straight line

a curve line

Solution

Question 6

$$\pi/4$$

$$\pi/2$$

$$\pi$$

$$2\pi$$

Solution

Question 7

SHM with period $$\pi$$ second

SHM with period $$2 \pi$$ second

oscillatory but not SHM

SHM with period $$\frac{1}{2\pi}$$ second

Question 8

$$\sqrt{\frac{K}{m}}$$

$$\sqrt{\frac{m}{K}}$$

$$\left ( \frac{K}{m} \right )^{2}$$

$$\left ( \frac{m}{K} \right )^{2}$$

Solution

Question 9

rotatory

oscillatory

curvilinear

rectilinear

Solution

Question 10

periodic but not oscillatory.

Periodic and oscillatory.

oscillatory but not periodic

neither periodic nor oscillatory

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.