Question

If two pendulums with time periods T1 and T2 have lengths 1.0 m and 4.0 m respectively, what is the ratio
?
T1 over T2​

Answer 1

Answer:

The answer tk your question is given and explained in the attachment.

Answer 2

Answer

• Ratio of T₁ over T₂  is , T₁ : T₂ =  1 : 2

Given

• Two pendulums with time periods T₁ and T₂  have lengths 1 m and 4 m respectively

To Find

• Ratio of T₁ and T₂

Concept Used

For a simple pendulum , Time period is ,

$\bullet \;\;\; \bf T=2 \pi \sqrt{\dfrac{l}{g}}$

Solution

Case - 1 : Pendulum 1

Time period = T₁

Length , l₁ = 1 m

Apply Time period for simple pendulum .

$\implies \rm T=2 \pi \sqrt{\dfrac{l}{g}}\\\\\implies \rm T_1=2 \pi \sqrt{\dfrac{1}{g}}\\\\\implies \rm T_1=\dfrac{2 \pi}{\sqrt{g}}$

Case - 2 : Pendulum 2

Time period = T₂

Length , l₂ = 4 m

Again apply Time period for simple pendulum .

$\implies \rm T=2 \pi \sqrt{\dfrac{l}{g}}\\\\\implies \rm T_2=2 \pi \sqrt{\dfrac{4}{g}}\\\\\implies \rm T_2=2 \pi \times \dfrac{2}{\sqrt{g}}\\\\\implies \rm T_2=\dfrac{4 \pi}{\sqrt{g}}$

Now , We need to find T₁ : T₂

$\implies \rm \dfrac{T_1}{T_2}\\\\\implies \rm \dfrac{\frac{2\pi}{\sqrt{g}}}{\frac{4\pi}{\sqrt{g}}}\\\\\implies \rm \dfrac{2\pi}{\cancel{\sqrt{g}}} \times \dfrac{\cancel{\sqrt{g}}}{4\pi}\\\\\implies \rm \dfrac{2\cancel{\pi}}{4\cancel{\pi}}\\\\\implies \rm \dfrac{2}{4}\\\\\implies \rm \dfrac{1}{2}$

So , T₁ : T₂  is  1 : 2