Tag: motion of a mass suspended by two springs

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

A bob of mass  $\mathrm { M }$  is hung using a string of length  $\mathrm { l }.$  A mass  $m$  moving with a velocity  $u$  pierces through the bob and emerges out with velocity  $\dfrac { u } { 3 } ,$  The frequency of oscillation of the bob considering as amplitude  $A$ is

  1. $2 \pi \sqrt { \dfrac { 3 m u } { 2 M A } }$
  2. $\dfrac { 1 } { 2 \pi } \sqrt { \dfrac { 2 m } { 3 M A } }$
  3. $\dfrac { 1 } { 2 \pi } \left( \dfrac { 2 m u } { 3 M A } \right)$

  4. cannot be found
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

A  body of mass 0.98 Kg is suspended from a spring of spring constant K = 2N/m. Then the period is. 

  1. 4.9s

  2. 4.4s

  3. 5.2s

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The period of a mass-spring system is T = 2 * pi * sqrt(m/k). Given m = 0.98 kg and k = 2 N/m, T = 2 * pi * sqrt(0.98/2) = 2 * pi * sqrt(0.49) = 2 * pi * 0.7 = 1.4 * pi. Using pi approx 3.14, T is approx 4.396s, which rounds to 4.4s.

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

Two particles  $A$  and  $B$  of equal masses are suspended from two massless springs of spring constants  $k _ { 1 }$  and  $k _ { 2 }$  respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitudes of  $A$  and  $B$  is

  1. $\sqrt { k _ { 1 } / k _ { 2 } }$
  2. $k _ { 1 } / k _ { 2 }$
  3. $\sqrt { k _ { 2 } / k _ { 1 } }$
  4. $k _ { 2 } / k _ { 1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Maximum velocity v_max = A * omega = A * sqrt(k/m). Since masses are equal, v_max is proportional to A * sqrt(k). For equal velocities, A1 * sqrt(k1) = A2 * sqrt(k2), so A1/A2 = sqrt(k2/k1).

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

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. $7.1\, cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The period T = 2 * pi * sqrt(m/k). Given T = 0.5s and m = 4kg, 0.5 = 2 * pi * sqrt(4/k). Squaring both sides: 0.25 = 4 * pi^2 * (4/k), so k = 64 * pi^2. The extension x = mg/k = (4 * 9.8) / (64 * pi^2) approx 39.2 / 631.65 approx 0.06206 m, which is 6.2 cm.

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

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

For springs in series, the effective spring constant k_eff is given by 1/k_eff = 1/k1 + 1/k2, which simplifies to k_eff = (k1 * k2) / (k1 + k2). The time period T = 2 * pi * sqrt(m/k_eff) = 2 * pi * sqrt(m * (k1 + k2) / (k1 * k2)).

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

A block of mass m is suspended separately by two different spring have time period $ t _1 and t _2 $ . if same mass is connected to parallel combination of both springs , then its time period is given by

  1. $ \dfrac {t _1t _2}{t _1 +t _2} $
  2. $ \dfrac {t _1t _2}{\sqrt {t^2 _1+ t^2 _1} } $
  3. $ \sqrt {\dfrac { t _1t _2}{ t _1 +t _2}} $
  4. $\sqrt {(t _1)^2 + (t _2)^2} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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

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}}$