6. A block with a mass M is attached to a vertical spring with a spring constant k. When the block is displaced from equilibrium and released its period is T. A second identical spring k is added to the first spring in parallel.What is the period of oscillations when the block is suspended from two springs? A. 2T B. sqrt (2)T C.T D. (T)/(sqrt (2))

To solve this problem, we need to understand how the period of oscillation of a spring-block system is affected by the spring constant.<br /><br />The period of oscillation \( T \) for a block-spring system is given by the formula:<br /><br />\[ T = 2\pi \sqrt{\frac{M}{k}} \]<br /><br />where:<br />- \( M \) is the mass of the block,<br />- \( k \) is the spring constant.<br /><br />When two identical springs are added in parallel, the effective spring constant \( k_{\text{eff}} \) is given by:<br /><br />\[ k_{\text{eff}} = k_1 + k_2 = k + k = 2k \]<br /><br />where \( k_1 \) and \( k_2 \) are the spring constants of the two springs.<br /><br />Now, we need to find the new period \( T_{\text{new}} \) with the effective spring constant \( k_{\text{eff}} \):<br /><br />\[ T_{\text{new}} = 2\pi \sqrt{\frac{M}{k_{\text{eff}}}} = 2\pi \sqrt{\frac{M}{2k}} \]<br /><br />We can express this in terms of the original period \( T \):<br /><br />\[ T_{\text{new}} = 2\pi \sqrt{\frac{M}{2k}} = 2\pi \sqrt{\frac{1}{2} \cdot \frac{M}{k}} = 2\pi \cdot \frac{1}{\sqrt{2}} \cdot \sqrt{\frac{M}{k}} = \frac{1}{\sqrt{2}} \cdot 2\pi \sqrt{\frac{M}{k}} = \frac{T}{\sqrt{2}} \]<br /><br />Therefore, the period of oscillations when the block is suspended from two springs is:<br /><br />\[ \boxed{\frac{T}{\sqrt{2}}} \]<br /><br />So the correct answer is:<br /><br />D. \(\frac{T}{\sqrt{2}}\)