Tag: oscillatory motion

Questions Related to oscillatory motion

A force of 6.4 N stretches a vertical spring by 0.1 m. The mass that must be suspended from the spring so that it oscillates with a period of ($\pi/4$) sec is:  

  1. $(\pi/4)$ kg

  2. 1 kg

  3. $(1 / \pi)$

  4. 10 kg


Correct Option: B
Explanation:

$\begin{array}{l} k=\frac { f }{ x } =\frac { { 6.4 } }{ { 0.1 } } =64 \ T=2\pi \sqrt { \frac { m }{ k }  }  \ \frac { \pi  }{ 4 } =2\pi \sqrt { \frac { m }{ { 64 } }  }  \ m=1\, kg \end{array}$

Hence,
option $(B)$ is correct answer.

A spring of force constant $800 Nm^{-1}$ has an extension of 5 cm . The work done in extending it from 5 cm to 15 cm is

  1. 16 J

  2. 8 J

  3. 32 J

  4. 24 J


Correct Option: B

A spring of spring constant ($k$) is attached to a block of mass ($m$). During free fall its time period of oscillations will be

  1. Zero

  2. Infinite

  3. $2\pi \sqrt{\cfrac{m}{k}}$

  4. $\pi \sqrt{\cfrac{m}{k}}$


Correct Option: C

Two identical springs are attached to a mass and the system is made to oscillate. ${ T } _{ 1 }$ is the time period when springs are joined in parallel and ${ T } _{ 2 }$ is the time period when they are joined in series then

  1. ${ T } _{ 1 }=2{ T } _{ 2 }$

  2. ${ T } _{ 1 }=\sqrt { 2 } { T } _{ 2 }$

  3. ${ T } _{ 2 }=2{ T } _{ 1 }$

  4. ${ T } _{ 2 }=\sqrt { 2 } { T } _{ 1 }$


Correct Option: C

A block tied between two springs is in equilibrium. If upper spring is cut then the acceleration of the block just after cut is 6 ${ m/s }^{ 2 }$ downwards. Now, if instead of upper spring, lower spring is cut then the magnitude of acceleration of the block just after the cut will be : (Take g = 10 ${ m/s }^{ 2 }$)

  1. 16 ${ m/s }^{ 2 }$

  2. 4 ${ m/s }^{ 2 }$

  3. Cannot be determined

  4. None of these


Correct Option: B

A light spring of length 20 cm and force constant 2 N/cm is placed vertically on a table. A small block of mass 1 kg falls on it. The length h from the surface of the table at which the block will have the maximum velocity is  

  1. 20 cm

  2. 15 cm

  3. 10 cm

  4. 5 cm


Correct Option: D

Two dissimilar spring fixed at one end are stretched by 10cm and 20cm respectively, when masses ${ m } _{ 1 }$ and ${ m } _{ 2 }$ are suspended at their lower ends. When displaced slightly from their mean positions and released, they will oscillate with period in the ratio

  1. 1 : 2

  2. 2 : 1

  3. 1 : 1.41

  4. 1.41 :4


Correct Option: A

A body of mass $4\, kg$ hangs from a spring and oscillates with a period $0.5$ second. On the removed of the body, the spring is shortened by

  1. $6.4\, cm$

  2. $6.2\, cm$

  3. $6.8\, cm$

  4. $7.1\, cm$


Correct Option: B

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are $ K _1 and K _2 $. The time period of the suspended mass will be-

  1. $ T = 2 \pi \sqrt { \left( \dfrac { m }{ k _ 1-k _ 2 } \right) } $

  2. $ T = 2 \pi \sqrt { \left( \dfrac { m }{ k _ 1+k _ 2 } \right) } $

  3. $ T = 2 \pi \sqrt { \left( \dfrac { m\left( k _ 1+k _ 2 \right) }{ k _{ 1 }k _{ 2 } } \right) } $

  4. $ T = 2 \pi \sqrt { \left( \dfrac { mk _ 1k _ 2 }{ k _{ 1 }+k _{ 2 } } \right) } $


Correct Option: C

Two massless springs of force constants ${ k } _{ 1 }$ and ${ k } _{ 2 }$ are joined end to end. The resultant force constant $k$ of the system is

  1. $k=\dfrac { { k } _{ 1 }+{ k } _{ 2 } }{ { k } _{ 1 }{ k } _{ 2 } } $

  2. $k=\dfrac { { k } _{ 1 }-{ k } _{ 2 } }{ { k } _{ 1 }{ k } _{ 2 } } $

  3. $k=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }+{ k } _{ 2 } } $

  4. $k=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }-{ k } _{ 2 } } $


Correct Option: C
Explanation:

In series, resultant force constant is given as
  $\dfrac { 1 }{ { k } _{ eq } } =\dfrac { 1 }{ { k } _{ 1 } } +\dfrac { 1 }{ { k } _{ 2 } } $
$\Rightarrow { k } _{ eq }=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }+{ k } _{ 2 } } $