A certain spring elongates 9 mm when it is suspended vertically and a block of mass M is hung on it. The natural frequency of this mass-spring system is:
A) is 0.088 rad/s
B) is 33 rad/s
C) is 200 rad/s
D) is 1140 rad/s
E) cannot be computed unless the value of M is given

Question

Quizplus · Accepted Answer

The natural frequency of a mass-spring system is given by:
w = sqrt(k/m)

where k is the spring constant and m is the mass.

Since the spring elongates 9 mm (or 0.009 m) when a mass M is hung on it, we can use Hooke's law to find the spring constant:

F = kx
Mg = kx
k = Mg/x

where g is the acceleration due to gravity (9.81 m/s^2).

Substituting k into the expression for the natural frequency, we get:

w = sqrt((Mg/x)/m)
w = sqrt((Mg)/(mx))

Since we are not given the value of M, we cannot compute the natural frequency exactly. However, we can make an estimate based on typical values of spring constants and masses. For example, if we assume that the mass is 0.1 kg and the spring constant is 100 N/m (which are reasonable values for a typical spring), then we get:

w = sqrt((0.1 kg)(9.81 m/s^2)/(0.009 m)(100 N/m))
w = sqrt(109)
w = 10.44 rad/s

Therefore, the best choice is B, which is closest to our estimate of 10.44 rad/s.

A Certain Spring Elongates 9 Mm When It Is Suspended