A spring of force constant \(k\) is cut into lengths of ratio \(1:2:3\). They are connected in series and the new force constant is \(k'\). Then they are connected in parallel and force constant is \(k''\). Then \(k':k''\) is:
1. \(1:9\)
2. \(1:11\)
3. \(1:14\)
4. \(1:6\)

Subtopic:  Combination of Springs |
 71%
From NCERT
NEET - 2017
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When a mass \(m\) is connected individually to two springs \(S_1\) and \(S_2,\) the oscillation frequencies are \(\nu_1\) and \(\nu_2.\) If the same mass is attached to the two springs as shown in the figure, the oscillation frequency would be: 

         

1. \(v_2+v_2\) 2. \(\sqrt{v_1^2+v_2^2}\)
3. \(\left(\dfrac{1}{v_1}+\dfrac{1}{v_1}\right)^{-1}\) 4. \(\sqrt{v_1^2-v_2^2}\)
Subtopic:  Combination of Springs |
 72%
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A mass of \(2.0\) kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released, the mass executes a simple harmonic motion. The spring constant is \(200\) N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? 
(Take \(g=10\) m/s2
                

1. \(8.0\) cm
2. \(10.0\) cm
3. any value less than \(12.0\) cm
4. \(4.0\) cm
Subtopic:  Combination of Springs |
 78%
From NCERT
AIPMT - 2007
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When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is \(t _1\) and \(t_2\) respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes \(t_0.\) The correct relation between \(t_0,\) \(t_1\) & \(t_2\) is:

1. t02=t12+t22

2. t0-2=t1-2+t2-2

3. t0-1=t1-1+t2-1

4. t0=t1+t2

Subtopic:  Combination of Springs |
 69%
From NCERT
AIPMT - 2002
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