Question

simple pendulum is suspended under gravity oscillates in vertical plane with time period 2 s. The bob is given charge q and an electric field E is switched on in horizontal direction. If qE = 3mg, then new time period in seconds is ______(Given, 103/4 = 5.62) Answer:

Answer 1

Answer:

T=2π

a

L

Now new force =mg+qE

Manet=mg+qE

anet=

M

mg+qE

∴anet=g+

M

qE

So, T=2π

q+

M

qE

l

Explanation:

I hope it helps u

Answer 2

Answer:

The answer is 1 sec.

Explanation:

For any oscillating pendulum, the time period is defined as:

$T = \frac{2\pi }{\sqrt[]{a} }$

Where,

T = Time period

a = effective acceleration

Case I: When the simple pendulum oscillates only under gravity.

Acceleration a = g

Hence,

$T = \frac{2\pi }{\sqrt[]{g} }$

Given, T = 2 s

Hence,    $\frac{2\pi }{\sqrt[]{g} }$ = 2 s

Case II: When the simple pendulum oscillates under gravity and electric field.

Total force = gravitational force + electric force

Hence,

ma = mg + qE

Given, qE = 3mg

Hence, ma = mg + 3mg = 4mg

Hence, a = 4g

New time period:

$T = \frac{2\pi }{\sqrt[]{a} }$

$T = \frac{2\pi }{\sqrt[]{4g} }$

$T =\frac{1}{2} \frac{2\pi }{\sqrt[]{g} }$

$T = \frac{1}{2} * 2$

T = 1 s

Hence, new time period will be 1 sec.

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