Question

A spring of constant k is cut into two parts having the ratio of their length 3:4 and now these parts having spring constants k1 and k2 are connected across a block of mass m. The time period of SHM of block will be

Answer 1

under root (m/7k)
this may help u..............

Answer 2

Answer:

The time of SHM of the given block is $T=2\pi \sqrt{\frac{12m}{49K } } s$

Explanation:

When a spring is cut its length(L) and spring constant(K) changes but their product remains constant i.e.

                                          $KL=Constant$

Since spring is cut in the ratio of its length as 3:4,

$KL=K_{1}\frac{3L}{7}=K_{2}\frac{4L}{7}$

On solving we get,

$K_{1}=\frac{7K}{3} \\K_{2}=\frac{7K}{4}$

As both the springs are now connected across the block their equivalent spring constant will change to,

$K_{eq}= K_{1}+ K_{2}\\K_{eq}=\frac{7K}{3}+ \frac{7K}{4} \\K_{eq}=\frac{49K}{12}$

Time period of the given block of mass m will be,

$T=2\pi \sqrt{\frac{m}{K_{eq} } }\\T=2\pi \sqrt{\frac{12m}{49K } }$

Hence, The time of SHM of the given block is $T=2\pi \sqrt{\frac{12m}{49K } } s$