Tag: oscillatory motion

Questions Related to oscillatory motion

One end of a long metallic wire of length $L$ area of cross-section $A$ and Young's modulus $Y$ is tied to the ceiling. The other end is tied to a massless spring of force constant $k$. A mass $m$ hangs freely from the free end of the spring. It is slightly pulled down and released. Its time period is given by-

  1. $\displaystyle 2\pi \sqrt{\frac{m}{k}}$

  2. $\displaystyle 2\pi \sqrt{\frac{mYA}{kL}}$

  3. $\displaystyle 2\pi \sqrt{\frac{mk}{YA}}$

  4. $\displaystyle 2\pi \sqrt{\frac{m(kL+YA)}{kYA}}$


Correct Option: D
Explanation:
$F = \dfrac{YA\Delta l}{L} = k _2 \Delta l$
we can consider the system as two springs in series hence 
$\dfrac{1}{k _{eq}} = \dfrac{1}{k _1} +\dfrac{1}{k _2}$
$=\dfrac{1}{k} + \dfrac{L}{YA} = \dfrac{YAk +kL}{YAk}$

$ T = 2\pi \sqrt{\dfrac{m}{k _{eq}}} = 2\pi \sqrt{\dfrac{m(YAk + kL)}{YAk}}$

The frequency $f$ of vibrations of a mass $m$ suspended from a spring of spring constant $k$ is given by $f = Cm^xk^y$, where $C$ is a dimensionless constant. The values of $x$ and $y$ are respectively:

  1. $\dfrac{1}{2}, \dfrac{1}{2}$

  2. $-\dfrac{1}{2}, -\dfrac{1}{2}$

  3. $\dfrac{1}{2}, -\dfrac{1}{2}$

  4. $-\dfrac{1}{2}, \dfrac{1}{2}$


Correct Option: D
Explanation:
We know $F=-KK\Rightarrow dim\left( K \right) =\left[ { MLT }^{ -2 } \right] \left[ { L }^{ -1 } \right] ={ ML }^{ 0 }{ T }^{ -2 }$
$dim\left( M \right) ={ ML }^{ 0 }{ T }^{ 0 }$    $dim\left( f \right) =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 } \right] $
$f={ Cm }^{ x }{ K }^{ y }\Rightarrow { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 }={ \left[ { ML }^{ 0 }{ T }^{ 0 } \right]  }^{ k }{ \left[ { ML }^{ 0 }{ T }^{ -2 } \right]  }^{ y }$
$\Rightarrow$  Comparing powers of $M,L$ and $T$ gives,
$x+y=0\quad \quad -2y=-1\quad \Rightarrow \quad y=\dfrac { 1 }{ 2 } $
and $x=-1/2$

A uniform spring has certain mass suspended from it and it's period of vertical oscillations is ${t} _{1}$. The spring is now cut in $2$ parts having lengths in ratio $1:2$  and these springs are now connected in series and then in parallel. find out the ratio of the time period of these two ossillation?

  1. $1$

  2. $\sin \theta$

  3. $\sqrt {\dfrac {2}{9}}$

  4. $\sqrt {\dfrac {9}{2}}$


Correct Option: C
Explanation:
Let $k$ be initial force constant of spring,${k} _{1}$ and ${k} _{2}$ be the force constant of neew springs
We can derive,
$ kl= constant $
$\Rightarrow \dfrac{{x} _{1}}{{x} _{2}}=1/2$
$\Rightarrow \dfrac{{k} _{1}}{{k} _{2}}=2$     ........$(1)$
$so k _1=3k, k _2=3k/2 $
As initially these lengths were in series:
$\dfrac{1}{k}=\dfrac{1}{{k} _{1}}+\dfrac{1}{{k} _{2}}$
$\Rightarrow \dfrac{1}{k' _1}=\dfrac{1}{3{k} _{1}}+\dfrac{2}{3{k} _{1}}$
$\Rightarrow \dfrac{1}{k' _1}=\dfrac{1}{{k} _{1}}$
$\Rightarrow {k'} _{1}= k\ $
When these two stringd are connected in parallel,
${k _2}^{\prime}={k} _{1}+{k} _{2}$

${k _2}^{\prime}=\dfrac{3k}{2}+3k$

${k _2}^{\prime}=\dfrac{9k}{2}$

Time period is 
$\dfrac{{T _1}^{\prime}}{T' _2}=\sqrt{\dfrac{k' _1}{{k' _2}^{\prime}}}$
$\Rightarrow \dfrac{{T _1}^{\prime}}{T' _2}=\sqrt{\dfrac{2}{9}}$

An infinite number of springs having force constants as K, 2K, 4K, 8K, .......$\displaystyle \infty $ respectively are connected in series; then equivalent spring constant is 

  1. K

  2. 2K

  3. $\displaystyle \frac{K}{2}$

  4. $\displaystyle \infty $


Correct Option: C
Explanation:

For the springs connected in series

$\dfrac{1}{K _{eq}}=\dfrac{1}{K}+\dfrac{1}{2K}+\dfrac{1}{4K}+\dfrac{1}{8K}+......$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(1+\dfrac{1}{2}+\dfrac{1}{4}+.....)$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(\dfrac{1}{1-\dfrac{1}{2}})$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(2)$
$K _{eq}=\dfrac{K}{2}$

A large box is accelerated up the inclined plane with an acceleration a and pendulum is kept vertical (Somehow by an external agent) as shown in figure.Now if the pendulum is set free to oscillate from such position, then what is the tension in the string immediately after the pendulum is set free? (mass of $500m$)

  1. $mg$

  2. $ma _{o} \sin\theta$

  3. $\left( m g + m a _ { 0 } \sin \theta \right)$

  4. Zero


Correct Option: C

 The time period of oscillation of a torsional pendulum of moment of inertia I is

  1. $T =2 \pi \sqrt{I/k}$

  2. $T =2 \pi \sqrt{I/2k}$

  3. $T =2 \pi \sqrt{2I/k}$

  4. $T =2 \pi \sqrt{I/4k}$


Correct Option: A
Explanation:

The time period of oscillations of a torsional pendulum is $T =2 \pi \sqrt(I/k)$

The correct option is (a)

A bullet of mass $'m'$ hits a pendulum bob of mass $'2m'$ with a velocity $'v'$ and comes out of the bob with velocity $v/2$. Length of the pendulum is $2$ meter and $g=10 ms^{-2}$. The minimum value of $'v'$ for the bullet so that the bob may complete one revolution in the verticle is

  1. $40 ms^{-1}$

  2. $2.20 ms^{-1}$

  3. $3.15 ms^{-1}$

  4. $10 ms^{-1}$


Correct Option: A

Time period of a disc about a tangent parallel to the diameter is same as the time period of a simple pendulum. The ratio of radius of disc to the length of pendulum is :

  1. $\dfrac { 1 } { 4 }$

  2. $\dfrac { 4 } { 5 }$

  3. $\dfrac { 2 } { 3 }$

  4. $\dfrac { 1 } { 2 }$


Correct Option: C

A pendulum of mass $m$ hangs from a support fixed to a trolley. The direction of the string (i.e.., angle $\theta$) when the trolley rolls up a plane of inclination $\alpha$ with acceleration $'a'$ is

  1. Zero

  2. $\tan^{-1} \alpha$

  3. $\tan^{-1}\dfrac{a+g \sin \alpha}{g \cos \alpha}$

  4. $\tan^{-1}\dfrac{a}{g}$


Correct Option: C

The oscillations of a pendulum slow down due to

  1. the force exerted by air and friction at the support

  2. the foce exerted by air only

  3. the forces exerted by friction at the support only

  4. none of these


Correct Option: A
Explanation:

The pendulum slows down due to friction force exerted by air and friction at the support.