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OpenStudy (anonymous):
For which value of instantaneous displacement, the K.E and P.E are equal ??? (x. = amplitude) A. x. /4 B. 3x. /4 C. x. /2 D. x. /undr root 2
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OpenStudy (anonymous):
a x/4
OpenStudy (anonymous):
@keshavsarda how?
OpenStudy (anonymous):
this is realeted to shm
OpenStudy (anonymous):
@keshavsarda please explain
OpenStudy (anonymous):
KE=1/2*m*omega^2*v^2 PE=1/2*m*v^2-KE
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