Question

a body of 0.25 kg execute shm given by=y=0.4 sin 0.5πt m calculate the amplitude angular frequency maximum velocity and maximum acceleration​

Answer 1

Answer

v max =ωA=2πfA = (2π)(2)(8×10 −3 )

=0.101 m/s

a max =ω

2A=(2πf)

2A=4×π

2×4×8×10 −3

=1.264 m/s

2F max=ma

max=(0.5)(1.264)=0.632 N

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