A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released. Assuming no friction and no external force, the natural frequency W (measured in radians per unit time) for the system is? (Recall that the acceleration due to gravity is 32ft/sec2). a) None of the other alternatives is correct.
b) W = v2 3
c)w=212
d) w = 4/6
e) w=213

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

substituting the values of the variables into the equation, we have

k = mg/x  

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

i think the answer is a, hope this : )