On a smooth inclined plane, a body of mass \(M\) is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant \(K\), the period of oscillation of the body (assuming the springs as massless) will be:
                
1. \(2\pi \left( \frac{M}{2K}\right)^{\frac{1}{2}}\)
2. \(2\pi \left( \frac{2M}{K}\right)^{\frac{1}{2}}\)
3. \(2\pi \left(\frac{Mgsin\theta}{2K}\right)\)
4. \(2\pi \left( \frac{2Mg}{K}\right)^{\frac{1}{2}}\)

Subtopic:  Combination of Springs |
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A mass m is suspended from two springs of spring constant k1 and k2 as shown in the figure below. The time period of vertical oscillations of the mass will be
                

1. 2πk1+k2m

2. 2πmk1+k2

3. 2πm(k1k2)k1+k2

4. 2πmk1+k2k1k2

Subtopic:  Combination of Springs |
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From NCERT
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All the surfaces are smooth and the system, given below, is oscillating with an amplitude \(\mathrm{A}.\) What is the extension of spring having spring constant \(\mathrm{k_1},\) when the block is at the extreme position?
             

1. \({k_1 \over k_1+k_2} \text{A}\) 2. \({k_2A \over k_1+k_2}\)
3. \(\mathrm{A}\) 4. \(\text{A} \over 2\)
Subtopic:  Combination of Springs |
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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 |
 68%
From NCERT
AIPMT - 2002
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