Oscillations Test

Result

Oscillations Test - 42

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

A simple pendulum has time period $$T _ { 1 }.$$ The point of suspension is now moved upward according to equation $$y = K t ^ { 2 }$$ where $$K = 2.5 \mathrm { m } / \mathrm { s } ^ { 2 }.$$ If new time period is $$T _ { 2 }$$ then ratio will be

A
$$2 / 3$$

B
$$5 / 6$$

C
$$6 / 5$$

D
$$3 / 2$$

Solution

Given,

Displacement,

$$ y=k{{t}^{2}} $$

$$ \dfrac{{{d}^{2}}y}{d{{t}^{2}}}=2k $$

Since, k=1m/s²

$$ \dfrac{{{d}^{2}}y}{d{{t}^{2}}}={{a}_{y}}=2\,m/{{s}^{2}} $$

$$ {{T}_{1}}=2\pi \sqrt{\dfrac{l}{g}}.........(1) $$

$$ and $$

$$ {{T}_{2}}=2\pi \sqrt{\dfrac{l}{g+{{a}_{y}}}}.........(2) $$

Taking ratios of equation (1) and (2)

$$\dfrac{{{T}^{2}_{1}}}{{{T}^{2}_{2}}}=\dfrac{g+{{a}_{y}}}{g}=\dfrac{10+2}{10}=\dfrac{6}{5}$$
Question 2
1 / -0

The time period of oscillation of a $$SHM$$ is $$\frac { \pi } { 2 } s.$$ Its acceleration at a phase angle $$\frac { \pi } { 3 }\ rad$$ from extreme position is $$2 m s ^ { - 2 }.$$ What is its velocity at a displacement equal to half of its amplitude from mean position? (in $$m s ^ { - 1 }$$)

A
$$0.707$$

B
$$0.866$$

C
$$\sqrt { 2 }$$

D
$$\sqrt { 3 }$$
Question 3
1 / -0

A particle undergoes SHM. When its displacement is 8 cm, it has speed 3 m/s and a speed 4 m/s at displacement of 6cm. Find time period of oscillation.

A
$$4\pi$$

B
$$6\pi$$

C
$$7\pi$$

D
$$8\pi$$

Solution

using the relation
$$v=w\sqrt{A^2-x^2}$$
from given ,
$$(i)$$ when $$x=8,v=3$$
$$3=w\sqrt{A^2-64}$$ $$\rightarrow (1)$$
$$(ii)$$ when $$x=6,v=4$$
$$4=w\sqrt{A^2-36}$$ $$\rightarrow (2)$$
dividing $$(1)\div(2)$$given
$$\left(\dfrac{3}{4}\right)^2\dfrac{A^2-64}{A^2-36}\Rightarrow A^2=\dfrac{16\times 64-9\times 36}{7}=100$$
$$\Rightarrow A=10\ cm$$
Putting $$A$$ in $$(1)$$ gives,
$$3=w\sqrt{10^2-b4}\Rightarrow w=\dfrac{3}{6}=\dfrac{1}{2}rad/s$$
$$\therefore $$ Time period of oscillation,
$$T=\dfrac{2\pi}{w}=4\pi$$
Question 4
1 / -0

If a spring has time $$T$$, and is cut into $$n$$ parts , then the time period of each part will be

A
$$T\sqrt{n}$$

B
$$\dfrac{T}{\sqrt{n}}$$

C
$$nT$$

D
$$T$$

Solution

Spring constant will becomes n times.
$$T\propto \dfrac{1}{\sqrt{k}}$$
$$T_{final}=\dfrac{T}{\sqrt{n}}$$
Question 5
1 / -0

If function $$\sin^{2}(\omega t)$$ represents position of a particle from the origin as function of time, then motion of particle is

A
A simple harmonic motion with a period $$2\pi/\omega$$

B
A simple harmonic motion with a period $$\pi/\omega$$

C
A period motion but not simple harmonic motion with a period $$2\pi/\omega$$

D
A period motion but not simple harmonic motion with a period $$\pi/\omega$$

Solution
Question 6
1 / -0

A block of mass $$10\,Kg$$ is suspended through two light spring balances as shown in figure.

A
Both the scales will read $$10\,Kg$$

B
Both the scales will read $$5\,Kg$$

C
The upper scale will read $$10\,Kg$$ and the lower zero

D
The readings may be anything but their sum will be $$10\,Kg$$

Solution

Since the block and the balances are in equilibrium and considering that balances are massless.

$$\Rightarrow T_1 = 10kg$$ and

$$T_2 = T_1$$

$$\Rightarrow T_ = T_2 = 10kg$$

and, hence, both the balances will read 10kg.
Question 7
1 / -0

A particle moves along a straight line such that its displacement(s) at any time(t) is given by $$S=t^3-3t^2+2$$. The velocity of the particle when its acceleration zero is?

A
$$-2$$ units

B
$$-8$$ units

C
$$0$$ units

D
$$-9$$ units

Solution

Displacement is given as$$=({ t }^{ 3 }−3{ t }^{ 2 }+2)\Longrightarrow (1)$$

We know that $$a=\dfrac { { d }^{ 2 }s }{ d{ t }^{ 2 } } $$

Differentiating (1 ) once with respect to t we get

$$\dfrac { ds }{ dt } =\dfrac { d }{ dt } ({ t }^{ 3 }−3{ t }^{ 2 }+2)=3{ t }^{ 2 }−6t$$

Differentiating once again we get

$$\dfrac { { d }^{ 2 }s }{ d{ t }^{ 2 } } =\dfrac { d }{ dt } (3{ t }^{ 2 }−6t)=6t−6$$

Given condition is $$\dfrac { { d }^{ 2 }s }{ d{ t }^{ 2 } } =6t−6=0$$

$$t=1s$$

$$s(1)=(13−3×12+2)=0m$$
Question 8
1 / -0

The time period of a particle in simple harmonic motion is equal to the time between consecutive apperance of the particle at a particular point in its motion. This point is

A
The mean position

B
An extreme position

C
Between the mean position and the positive extreme

D
Between the mean position and the negative extreme

Solution
Question 9
1 / -0

A particle is executing S.H.M. with amplitude $$'a'$$ and has maximum velocity $$'v'$$. Its speed at displacement $$a/2$$ will be

A
$$0.866\ v$$

B
$$v/2$$

C
$$v$$

D
$$v/4$$

Solution

So option A is correct.
Question 10
1 / -0

Spring in vehicles are introduces to:

A
Reduce

B
Reduce impluse

C
Reduce force

D
Reduce velocity

Solution