Oscillations Test

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Oscillations Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

A clock which has pendulum made of brass keeps correct time at $$30^\circ {\text{C}}$$. How many seconds it will lose or gain in a day if temperature falls to $$0^\circ {\text{C}}$$. $$\left( {\alpha _{brass} = 1.8 \times 10^{ - 3} /^\circ {\text{C}}^{ - 1} } \right)$$

A
$$23.33 sec$$

B
$$24.33 sec$$

C
$$26.33 sec$$

D
$$25.33 sec$$

Solution

$$T_0=2\pi\sqrt{\cfrac{L_0}{g}}\\T+2\pi\sqrt{\cfrac{L}{g}}\\T=T_0(1+\cfrac{\alpha\Delta\theta}{2})T=30[1+(1.8\times10^{-3}\times\cfrac{30}{2})]\\T=30[1+(0.9\times10^{-3}\times30)]\\T=23.33sec$$
Question 2
1 / -0

Calculate the period of oscillations of block of mass m attached with a set of springs as shown

A
2 $$\pi$$ $$\sqrt{m/3k}$$

B
2 $$\pi$$ $$\sqrt{3m/2k}$$

C
2 $$\pi$$ $$\sqrt{m/2k}$$

D
2 $$\pi$$ $$\sqrt{3k/m}$$

Solution

We know,

Time period$$(T)=2\pi\sqrt{\dfrac{m}{k_{eq}}}$$

Here,
$$k_{eq}=\dfrac{2k}{3}$$

Hence $$(T)=\sqrt{\dfrac{3m}{2k}}$$

Option $$\textbf B$$ is the correct answer.
Question 3
1 / -0

A thin spherical shell of mass $$M$$ and radius $$R$$ has a small hole. A particle of mass $$m$$ is released at its mouth. Then

A
the particle will execute SHM inside the shell

B
the particle will oscillate inside the shell, but the oscillations are not simple harmonic

C
the particle will not oscillate, but the speed of the particle will go on increasing

D
none of these

Solution

Firstly ,given that it is a thin shell i.e hollow sphere then it will perform oscillation on SHM.
In case of solid sphere it performs SHM.
Question 4
1 / -0

The string of a pendulum is horizontal initially. The mass of the bob attached is m. Now the string is released. 'R' is the length of the pendulum.
(i) Velocity of the string at an angle $$30^\circ$$ with the vertical is ________

A
$$\sqrt{\sqrt{2} gR}$$

B
$$\sqrt{\sqrt{3} gR}$$

C
$$\sqrt{\sqrt{5} gR}$$

D
$$\sqrt{\sqrt{7} gR}$$
Question 5
1 / -0

The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0 sec, the tension in the string when the bob passes through O is :

A
$$m (g + 2\pi^2h)$$

B
$$m (g + \pi^2h)$$

C
$$m (g + \frac{\pi^2}{2}h)$$

D
$$m (g + \frac{\pi^2}{3}h)$$

Solution

Tesnsion in the string when bob passes through lowest point
$$T=mg+\dfrac{mv^2}{r}=mg+mv \omega \ \ (\because v=r \omega)$$
putting $$v=\sqrt{2gh}$$ and $$ \omega =\dfrac{2 \pi}{T}=\dfrac{2 \pi}{2}=\pi$$
we get $$T=m(g+ \pi \sqrt{2gh})$$
Question 6
1 / -0

A bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of $$\sqrt{3 gl}.$$ Find the angle rotated by the string before it becomes slack.

A
$$\theta = cos^{-1} (-\dfrac{2}{3})$$

B
$$\theta = cos^{-1} (-\dfrac{3}{4})$$

C
$$\theta = cos^{-1} (-\dfrac{1}{3})$$

D
$$\theta = cos^{-1} (-\dfrac{5}{3})$$

Solution

$$V = \sqrt {3gl} $$
$$\dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2} = - mgh$$
$${v^2} = {u^2} - 2g\left( {1 + \cos \theta } \right)$$
$$ \Rightarrow {v^2} = 3gl - 2gl\left( {1 + \cos \theta } \right)$$--------------$$(1)$$
Again$$,$$
$$\dfrac{{m{v^2}}}{l} = mg\cos \theta $$
$${v^2} = \lg \cos \theta $$
from equation $$(1)$$ and $$(2),$$ we get
$$3gl - 2gl - 2gl\cos \theta = gl\cos \theta $$
$$3\cos \theta = 1 \Rightarrow \cos \theta = \frac{1}{3}$$
$$\theta = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)$$
So$$,$$ angle rotated before the string becomes slack
$$ = {180^0} - {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)$$
Hence,
option $$(C)$$ is correct answer.
Question 7
1 / -0

The velocity of a particle under going SHM at the mean position is $$2{ms}^{-1}$$. The velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude is

A
$$1.732m/s$$

B
$$\sqrt {3/2}m/s$$

C
$$1/2m/s$$

D
$$\cfrac{1}{\sqrt{3}}m/s$$

Solution

We know that for a particle undergoing SHM with an amplitude 'a', velocity 'v' and angular velocity '$$\omega $$'
$$v = \omega \sqrt {{a^2} - {x^2}} $$
At mean position x=a
And given v = 2m$${s^{ - 1}}$$ at mean position
$$\begin{array}{l} 2=\omega \sqrt { { a^{ 2 } } } =\omega a & -1 \\ v=\omega \sqrt { { a^{ 2 } }-\frac { { { a^{ 2 } } } }{ 4 } } =\omega \sqrt { \frac { { 3{ a^{ 2 } } } }{ 4 } } =\frac { { \sqrt { 3 } a\omega } }{ 2 } & -2 \\ Dividing, & \\ \frac { 2 }{ v } =\frac { { \omega a } }{ { \frac { { \omega a\sqrt { 3 } } }{ 2 } } } =\frac { 2 }{ { \sqrt { 3 } } } & \\ v=\sqrt { 3 } m{ s^{ -1 } }=1.732m{ s^{ -1 } } & \end{array}$$
Question 8
1 / -0

A particle of mass $$0.2\ kg$$ is executing $$SHM$$. The velocity displacement curve for which is shown in figure. The frequency of oscillation is:

A
$$\dfrac {1}{\pi}$$

B
$$\dfrac {1}{2\pi}$$

C
$$\dfrac {2}{\pi}$$

D
$$2\pi$$

Solution
Question 9
1 / -0

A body of mass $$m$$ has time period $$T_1$$ with one spring and has time period $$T_2$$ with another spring. if both the spring are connected in parallel and same mass is used, then new time period $$T$$ is given as

A
$$T^{2}= T_{1}^{2}+ T_{2}^{2}$$

B
$$T= T_{1}+ T_{2}$$

C
$$\dfrac{1}{T}=\dfrac{1}{ T_{1}}+\dfrac{1}{ T_{2}}$$

D
$$\dfrac{1}{ T^{2}}=\dfrac{1}{ T_{1}^{2}}+\dfrac{1}{ T_{2}^{2}}$$

Solution

Angular frequency, $$\omega =\sqrt{\dfrac{k}{m}}\ \ \ \Rightarrow \ \dfrac{2\pi }{T}=\sqrt{\dfrac{k}{m}}$$

$$\Rightarrow \ k=m{{\left( \dfrac{2\pi }{T} \right)}^{2}}$$ where, $$T$$ is time period.

Net Spring constant when two spring is connected in parallel.

$$ k={{k}_{1}}+{{k}_{2}} $$

$$\Rightarrow m{{\left( \dfrac{2\pi }{T} \right)}^{2}}=m{{\left( \dfrac{2\pi }{{{T}_{1}}} \right)}^{2}}+m{{\left( \dfrac{2\pi }{{{T}_{2}}} \right)}^{2}}$$

$$ \Rightarrow \dfrac{1}{{{T}^{2}}}=\dfrac{1}{{{T}_{1}}^{2}}+\dfrac{1}{{{T}_{2}}^{2}} $$
Question 10
1 / -0

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

A
3.14 m/s

B
6 m/s

C
1.54 m/s

D
8.26 m/s

Solution

P.E of pendulum is converted in to K.E and lost energy
$$P.E=K.E+lost energy\rightarrow1$$
given lost energy =10% of P.E=0.1P.E
P.E=K.E+0.1P.E
0.9P.E=K.E
$$0.9mgh=0.5mv^2$$
$$v^2=1.8gh$$
h=lenght of the pendulum=2
v = speed at lowest point
g=9.8
now $$v=5.94m/s$$

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Re-Attempt Weekly Quiz Competition

Oscillations Test - 37

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Oscillations Test - 37

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SHARING IS CARING

Question 1

$$23.33 sec$$

$$24.33 sec$$

$$26.33 sec$$

$$25.33 sec$$

Solution

Question 2

2 $$\pi$$ $$\sqrt{m/3k}$$

2 $$\pi$$ $$\sqrt{3m/2k}$$

2 $$\pi$$ $$\sqrt{m/2k}$$

2 $$\pi$$ $$\sqrt{3k/m}$$

Solution

Question 3

the particle will execute SHM inside the shell

the particle will oscillate inside the shell, but the oscillations are not simple harmonic

the particle will not oscillate, but the speed of the particle will go on increasing

none of these

Solution

Question 4

$$\sqrt{\sqrt{2} gR}$$

$$\sqrt{\sqrt{3} gR}$$

$$\sqrt{\sqrt{5} gR}$$

$$\sqrt{\sqrt{7} gR}$$

Question 5

$$m (g + 2\pi^2h)$$

$$m (g + \pi^2h)$$

$$m (g + \frac{\pi^2}{2}h)$$

$$m (g + \frac{\pi^2}{3}h)$$

Solution

Question 6

$$\theta = cos^{-1} (-\dfrac{2}{3})$$

$$\theta = cos^{-1} (-\dfrac{3}{4})$$

$$\theta = cos^{-1} (-\dfrac{1}{3})$$

$$\theta = cos^{-1} (-\dfrac{5}{3})$$

Solution

Question 7

$$1.732m/s$$

$$\sqrt {3/2}m/s$$

$$1/2m/s$$

$$\cfrac{1}{\sqrt{3}}m/s$$

Solution

Question 8

$$\dfrac {1}{\pi}$$

$$\dfrac {1}{2\pi}$$

$$\dfrac {2}{\pi}$$

$$2\pi$$

Solution

Question 9

$$T^{2}= T_{1}^{2}+ T_{2}^{2}$$

$$T= T_{1}+ T_{2}$$

$$\dfrac{1}{T}=\dfrac{1}{ T_{1}}+\dfrac{1}{ T_{2}}$$

$$\dfrac{1}{ T^{2}}=\dfrac{1}{ T_{1}^{2}}+\dfrac{1}{ T_{2}^{2}}$$

Solution

Question 10

3.14 m/s

6 m/s

1.54 m/s

8.26 m/s

Solution

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