Question

compare the period of:—(i) a simple pendulum at two places where G is 9.8 metre per second square and 2.48 metre per second square respectively. (ii) a simple pendulum of length 9m and 16 m at a place.




Please tell me fast please please

​

Answer 1

Answer:

Given:

1. Value of g is different as given
2. value of length is different as given

To find:

Comparison of time periods

Concept:

Whenever the question asks for compared, try to find out the Ratio .

Formulas used:

$t = 2\pi \sqrt{ \frac{l}{g} }$

, where t => time period, l => length and

g=> gravity

Calculation:

1st part :

Keeping length constant, we can say that

Time period ∝ (1/√g)

So Ratio of time period

$\frac{t1}{t2} = \frac{ \sqrt{g2} }{ \sqrt{g1} }$

$= > \frac{t1}{t2} = \sqrt{ \frac{9.8}{2.48} }$

$= > \frac{t1}{t2} = \sqrt{3.9516}$

$= > \frac{t1}{t2} = 1.987$

So the ratio is 1.987 : 1.

2nd part:

Keeping Gravitational acceleration constant, we can say that

Time period ∝ l

Ratio of time period :

$\frac{t1}{t2} = \frac{ \sqrt{l1} }{ \sqrt{l2} }$

$= > \frac{t1}{t2} = \frac{ \sqrt{9} }{ \sqrt{16} }$

$= > \frac{t1}{t2} = \frac{3}{4}$

So the required ratio is 3 : 4

Answer 2

Answer: 3:4

Explanation: