A uniform spring whose unstretched length is 1- has a force constant K- The spring is cut into two pieces of unstretched length l 1 and l 2- where l 2- l 1- n - n being an integer- Now a mass m is made to oscillate with first spring- The time period of its oscillation would beA- T-2 -m-Kn-1B- T-2 -mn-1-n KC- T-2 -m n-Kn-1D- T-2 -m-n K

The correct option is A Since k ∝ nbsp;1/l∴ nbsp; k_1/k_2 = l_2/l_1 = n ⇒ k_1 = nk_2 Now l = l_1 + l_2 ⇒1/k = 1/k_1 + 1/k_2 nbsp;⇒1/k = 1/k_1 + n/k_1 ⇒ ...

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Byju's Answer

Standard VII

Physics

Young's Modulus of Elasticity

A uniform spr...

Question

A uniform spring whose unstretched length is $l$ , has a force constant K. The spring is cut into two pieces of unstretched length $l_{1}$ and $l_{2}$ , where $\frac{l_{2}}{l_{1}} = n$ , n being an integer. Now a mass m is made to oscillate with first spring. The time period of its oscillation would be

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Solution

The correct option is C

Since $k \propto \frac{1}{l} ∴ \frac{k_{1}}{k_{2}} = \frac{l_{2}}{l_{1}} = n$

$\Rightarrow k_{1} = n k_{2}$

Now $l = l_{1} + l_{2} \Rightarrow \frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}} \Rightarrow \frac{1}{k} = \frac{1}{k_{1}} + \frac{n}{k_{1}}$

$\Rightarrow k_{1} = (n + 1) k ∴ T = 2 π \sqrt{\frac{m}{k_{1}}} = 2 π \sqrt{\frac{m}{(n + 1) k}}$

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A uniform spring whose unstretched length is l, has a force constant K. The spring is cut into two pieces of unstretched length l1 and l2 , where l2l1=n, n being an integer. Now a mass m is made to oscillate with first spring. The time period of its oscillation would be

The correct option is C Since k∝1l∴k1k2=l2l1=n ⇒k1=nk2 Now l=l1+l2⇒1k=1k1+1k2⇒1k=1k1+nk1 ⇒k1=(n+1)k∴T=2π√mk1=2π√m(n+1)k