A 5.0-kg block, initially at rest on a level, frictionless surface, is attached

Question

A 5.0-kg block, initially at rest on a level, frictionless surface, is attached to a Hooke's Law spring with spring constant k= 80 N/m. The spring, which is also level, is rigidly attached to a wall on the other end as shown in the diagram above. Also assume that there is no friction in the spring or at the point of attachment. The block is then streched from its equilibrium position at x= 0.00 m to a distance of 50 cm to the right and released. Taking the initial time to be t = 0.00s at the release point, answer the following:

What is the ordinary frequency in Hz? (1Hz= 1 cycle/s)

Explanation / Answer

Consider the point where it is stretched. i.e., at x=0.5m = 50 cm The forces acting in the horizontal direction are: Spring force = -K * x As, F = m*a = m * d2/dt2(x) => -K*x = m*d2/dt2(x) => K*x + m*d2/dt2(x) = 0 => (K/m) x + d2/dt2(x) = 0 Comparing this with standard form, (?2) x + d2/dt2(x) = 0 => ?2 = K/m = 80/5 =16 => ? = 4 But we also know that w = 2* pi * f => f= 4/ (2*pi) = 0.636 Hz

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