AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

(A)
At mean position,
The velocity of the body is maximum that is Aω
(B)
The total energy during SHM is,
TE=‌

1

2

kA2
The kinetic energy is,
KE=‌

1

2

k(A2−x2)
It is given that, KE=‌

3

4

TE
Substitute the values in the above equation.
‌

1

2

k(A2−x2)=‌

3

4

×‌

1

2

kA2
Simplify the above equation.
x=±‌

A

2

(C)
The total energy during SHM is,
TE=‌

1

2

kA2
The potential energy is,
PE=‌

1

2

kx2
It is given that, PE=‌

3

4

TE
Substitute the values in the above equation.
‌

1

2

kx2=‌

3

4

×‌

1

2

kA2
Simplify the above equation.
x=±‌

√3

2

A
(D)
The acceleration is maximum at extreme position.
Thus, option
(3) is correct.