TS EAMCET 18-July-2022 Shift 1 Solved Paper

The given situation is shown as

Let the angle be ωt when the KE is 75% of total energy.
Also, the moment of inertia of the bob of pendulum about the point of suspension
=I=ml2‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Also, velocity at that instant =lω‌cos‌ω‌t‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
Now, according to the question,
‌

1

2

mv2=‌

75

100

×‌

1

2

Iω2
where, (‌

1

2

Iω2) is the rotational KE of the particle,
‌

1

2

mv2=‌

75

100

×‌

1

2

×ml2×ω2‌‌‌ [from Eq. (i)] ‌
⇒‌‌‌

1

2

l2mω2cos2ωt=‌

3

4

×‌

1

2

×ml2ω2 [from Eq. (ii)
⇒‌‌cos2ωt=‌

3

4

⇒‌‌cos‌ω‌t=±‌

√3

2

⇒‌‌ωf=‌

π

6

(keeping only positive solution)
t=‌

π

6ω

=‌

πT

6×2π

(∴. Time period, T=‌

2

2

⇒‌‌t=‌

π×4

12π

(given, T=4s)
⇒‌‌t=‌

1

3

s