Question

time period of simple pendulum is measured to be equal to 4 s​

Answer 1

Answer:

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Answer 2

Answer:

The time period (t) the pendulum having shortest wire is option (a) 4/√7 seconds .

Explanation:

Let the initial length of pendulum wire = L meter

And Gravitational acceleration = g m/sec^2

= 9.8 m/sec^2

Provided that: Time period (T) = 4 sec

From Simple Harmonic Motions (S.H.M) we know that for a pendulam,

Time period (T):

$T = {2 \pi} \sqrt{ \frac{l}{g} } \: \: ...equation(i)$

From equation (i) we get the initial length of wire = L :

$L = \frac{ {T}^{2} g}{4 { \pi}^{2} } \\ L = \frac{ {4}^{2} \times 9.8}{ {4} \times { \pi}^{2} } \: meter \\ = 3.97179 \: meter$

Hence length of main wire = L = 3.97179 m

Now the string of pendulum wire is cut into three parts in ratio = 1 : 2 : 4

So the length of shortest wire = l

$l = (\frac{1}{1 + 2 + 4}) L \\ l = \frac{L}{7} \\ l= \frac{3.97179 }{7} m \\ = 0.5674 \: m$

Hence length of shortest pendulum wire l = 0.5674 m

So now time period (t) of this pendulum having shortest wire from equation (i) :

$T = {2 \pi} \sqrt{ \frac{l}{g} } \\ t = {2 \pi} \times (\sqrt{ \frac{0.5674}{9.8} } ) \: sec \\ t = 1.51185971 \: sec$

So we get:

$t = 1.51185971\: sec \\ ≈ \frac{4}{ \sqrt{7} } \: sec$

Hence the time period (t) the pendulum having shortest wire is 4/√7 seconds .