Question

A block is on a horizontal surface (a shake table) that is moving back and forth horizontally with simple harmonic motion of frequency 9.4 Hz. The coefficient of static friction between block and surface is 0.10. How great can the amplitude of the SHM be if the block is not to slip along the surface?

Answer 1

Solution The frequency is given by f = 9.4H2 & the coefficient of static friction ills ojo is to find the maximum distonice which is the amplitude xn. The mex fixed will result in the maximum acceleration. The displacement X(t) of a particle from its equalibitum position is described by The target X (H) = xm cos (wtto, Where xm = amplitude of Ships. Wit = Phase of the motion = phase constant. angular frequency W: ei We know that acceleration auf - Wexo at) = -w² km cos(wt+4). Fox amox GOS(++ ) = 1 4 Amaral - W²x mers. = mamai - maxm = mw²xmor the applied force dject is friction force

d reaches when the block is not maximum surface to slip along the fmon. els mg Now from eph (0) & (2), is my mw xman (93) Xmax. less (94) 0.10X (9.8 mis2) (anx 9.4 H2)2 Xmas 2.809 x 10 in 0.281 mm 2.1mm = 10²m) OR 0.0031 CM