Tag: oscillatory motion

Quizzes (7) Questions (124)

Questions Related to oscillatory motion

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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. $\displaystyle 2\pi \sqrt{\frac{m(kL+YA)}{kYA}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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}}$
Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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. $-\dfrac{1}{2}, \dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $x=-1/2$
Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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. $\sqrt {\dfrac {9}{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 
$\dfrac{{T _1}^{\prime}}{T' _2}=\sqrt{\dfrac{k' _1}{{k' _2}^{\prime}}}$
$\Rightarrow \dfrac{{T _1}^{\prime}}{T' _2}=\sqrt{\dfrac{2}{9}}$
Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the springs connected in series

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

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

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. $10 ms^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using conservation of momentum, the bob's velocity after impact is v/4. To complete a vertical circle, the bob must have a minimum velocity of sqrt(5gl) at the bottom, which is sqrt(5 * 10 * 2) = 10 m/s. Setting v/4 = 10 gives v = 40 m/s.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

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. $\tan^{-1}\dfrac{a}{g}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the non-inertial frame of the trolley, the bob experiences a pseudo-force 'ma' directed down the plane and gravity 'mg'. The effective acceleration vector is the sum of these components perpendicular and parallel to the plane, resulting in the angle tan(theta) = (a + g*sin(alpha)) / (g*cos(alpha)).