Question

find the position and time where kinetic become 3/4 of total energy​

Answer 1

Answer:

Text Solution

A

x=±A2

B

x=±A4

C

x=±3A−−−√2

D

x=±A2–√

Answer

A

Solution

K.E.=34T.E.

12mω2(A2−x2)=34×12mω2A2

A2−x2=34A2

A2−34A2=x2

A24=x2

∴x=A2

Explanation: