a body of mass 5 kg executes shm of amplitude of 0.5m.if the force constant is 100Nm calculate time period and maximum k.e and p.e
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ke= 1/2mv^2&pe=mgh apply this formulae
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Potential energy of the particle U =mass x potential= 0.1 x (5x2 + 10) Joule
Force F = – Du/dx= 0.1 (10x) =- x
But force = mass x acceleration .. F = 0.1 d2x /dt2
Hence equation of motion of the particle will be
0.1 d2x/dt2= -x or d2x/ dt2 +10x=0 or d2x /dt2+w2x =0
The above equation represents the SHM of the particle whose frequency of oscillation is
Force F = – Du/dx= 0.1 (10x) =- x
But force = mass x acceleration .. F = 0.1 d2x /dt2
Hence equation of motion of the particle will be
0.1 d2x/dt2= -x or d2x/ dt2 +10x=0 or d2x /dt2+w2x =0
The above equation represents the SHM of the particle whose frequency of oscillation is
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