A particle of mass 10 g is placed in- la- potential field given by V=50x2+100 J/kg. If the frequency of oscillation is found to be xπcycle/s, the value of x is _______.
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
answer is 5.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Given that potential energy is U=mV⇒ U=50x2+10010−2F=−dUdx=−(100x)10−2⇒ mω2x=−100×10−2x10×10−3ω2x=100×10−2x⇒ ω2=100⇒ ω=10rad/s⇒ f=ω2π=10π=5πHence x = 5.