Tag: horizontal oscillations of a mass attached to a spring

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Questions Related to horizontal oscillations of a mass attached to a spring

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A spring (of spring constant $= k )$ is cut into 4 equal parts and two parts are connected in parallel. What is the effective spring constant of these parts?

  1. $4K$
  2. $16K$
  3. $8K$
  4. $6K$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a spring is cut into 4 equal parts, each part has a spring constant k' = 4k. Connecting two of these in parallel gives an effective constant k_eff = k' + k' = 4k + 4k = 8k.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A spring of force constant k rests on asmooth floor, with one end fixed to awall. A block of mass $m$ hits the free end of the spring with velocity $v$ . Themaximum force exerted by the springon the wall is 

  1. $v \sqrt { \dfrac {m}{k} }$
  2. $m n \sqrt { k }$
  3. $m \sqrt { ( u k ) }$
  4. $k \sqrt { ( m v ) }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A body of mass 100 gm is suspended from a spring of force constant 50 N/m. The maximum acceleration produced in the spring is:

  1. g/2

  2. g

  3. g/3

  4. g/4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To calculate maximum acceleration consider spring at the equilibrium position: 
$F=ma$
where $F$ is force on spring which will be equal to $mg$ only (as gravity is the only force present. 
Therefore, $ma=mg$ or $a=g$.


Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Two springs mass systems having equal mass and spring constant ${k} _{1}$ and ${k} _{2}$. If the maximum velocities in two systems are equal then ratio of amplitude of 1st so that of 2nd is

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

Max velocity v_max = A * omega = A * sqrt(k/m). Since m is equal, v_max is proportional to A * sqrt(k). If v_max1 = v_max2, then A1 * sqrt(k1) = A2 * sqrt(k2). Thus, A1/A2 = sqrt(k2/k1).

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Two bodies A and B of equal mass are suspended from two spearte massless springs of spring contnat ${ K } _{ 1 }$ and  respectively If the bodies oscillate vertically such that their maximum velocities are ewqual , the ratio of the amplitude of A to that of B is 

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

This is identical to the previous question. v_max = A * sqrt(k/m). With equal masses, A1 * sqrt(K1) = A2 * sqrt(K2), so A1/A2 = sqrt(K2/K1).

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Three masses 700 gm, 500 gm and 400 gm are suspended at the end of the spring and they are in equilibrium. When the 700 gm mass is removed, the system oscillates with a period of 3 sec, when the 500 gm mass is also removed, it will oscillate with a period of 

  1. 1 sec

  2. 2 sec

  3. 3 sec
  4. $\sqrt { \dfrac { 12 }{ 5 } } sec$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

T = 2*pi*sqrt(m/k). Initially m = 0.7+0.5+0.4 = 1.6kg. When 0.7kg is removed, m = 0.9kg, T = 3s. So 3 = 2*pi*sqrt(0.9/k). When 0.5kg is also removed, m = 0.4kg. T' = 2*pi*sqrt(0.4/k). T'/3 = sqrt(0.4/0.9) = sqrt(4/9) = 2/3. T' = 3 * (2/3) = 2s.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A uniform spring of force constant $k$ is cut into pieces whose length are in the ratio $1 : 2$. What is the force constant of second piece in terms of $k$?

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

If a spring of constant k is cut into pieces of length L1 and L2, the constants are k1 = k * (L/L1) and k2 = k * (L/L2). For ratio 1:2, L1 = L/3 and L2 = 2L/3. The second piece (L2) has k2 = k * (L / (2L/3)) = 3k/2.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

The spring constant of a spring is $K$. When it is divided into n equal parts, then what is the spring constant of one piece :

  1. $nK$
  2. $K/n$
  3. $\dfrac{nK}{(n + 1)}$
  4. $\dfrac{(n +1) K}{n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a spring is divided into n equal parts, each part has a length L/n. Since k is inversely proportional to length, the new constant k' = k * (L / (L/n)) = n*k.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Two blocks m and m each of mass 3kg is connected with spring of constant 50 N/m. The coefficient of friction between m and ground is 0.4. The maximum amplitude of m during its oscillation, so that m does not move, is 

  1. 24 cm

  2. 12 cm

  3. 2.4 cm

  4. 6 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the block not to move, the maximum spring force must be less than or equal to the limiting friction. k*A <= mu*m*g. 50 * A <= 0.4 * 3 * 10. 50 * A <= 12. A <= 12/50 = 0.24 m = 24 cm.