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`        A body is oscillating simple harmonically with Tequal to 2 second. time that would it take for its kinetic energy o decrease from Kmax to  75% of Kmax is`
one year ago

Arun
24489 Points
```							 TE of an SHM is given as 1/2m(wA)^2K.E for any SHM can be written as 1/2 m(wA)^2 cos^2(wt). even if u dont know this dont worry, u can easily generalise this from the energy of spring-mass system.so for 75% KE,3/4 * 1/2 m(wA)^2=1/2 m(wA)^2 cos^2(wt)=> ((3)^1/2)/2=cos(wt)we get , wt = pi / 6w=2 pi/Tt=T/12 = 1/6 seconds. hope it helps RegardsArun
```
one year ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions