A mass-spring system oscillates with an amplitude of 5.00 cm. If the spring cons

Question

A mass-spring system oscillates with an amplitude of 5.00 cm. If the spring constant is 266 N/m and the mass is 457 g, determine the mechanical energy of the system.

Determine the maximum speed of the object.

Determine the maximum acceleration.

Explanation / Answer

spring has maximum amplitude of 5 cm. this is from the mean position at the mean position the forces on the mass are balanced. hence if x is the extension of spring in the mean position, kx=mg k is the spring constant ==> 266*x = 0.457*9.8 ==> x = 0.0168 m = 1.68 cm so the maximun extension in the spring = 1.68+5 = 6.68 cm now thr total mechanical energy can be found by (1/2)kx^2 where x is the maximum extension because at this point mass has no velocity. hence mechanical energy = (1/2)*266*(0.0668)^2 = 0.593 J the block has maximum speed when it is at the mean position and at mean position the energy in the spring is (1/2)*266*(0.0168)^2 so by conservation of energy we have, 0.593 = (1/2)mv^2 + (1/2)*266*(0.0168)^2 (1/2)*0.457*v^2 = 0.593 - 0.0375 = 0.5555 ==> v= sqrt(2*0.5555/0.457) = 1.56 m/s so maximum speed = 1.56 m/s the maximum accleration is proportional to the maximum displacement and it is at the extreme position. hence accleration a = kx/m = 266*0.05/0.457 = 29.10 m/s^2

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