Question

A 4.53-kg mass attached to the lower end of a spring whose upper end is fixed vibrates with natural period of 0.45 s. Determine the natural period when a 2.26-kg mass is attached to the midpoint of the same spring with the upper and lower ends fixed.

Answer 1

Answer:

$T_2=0.58s$

Step-by-step explanation:

As frequency is $\frac{1}{T}$ thus:

$f_1=1/T ; f_1=2.23Hz$

$w_1=\sqrt{k/m_1}$

$f_1=2\pi *w_1$

$w_1=f_1/2\pi$

$w_1=0.35Hz$

$k=w_1^2(m_1)$

$k=(0.35Kz)^2(4.53kg)$

$k=0.55 N/m$

$w_2=\sqrt{k/m_t}$

$m_t=m_1+m_2$

$w_2=\sqrt{(0.55 N/m)/(6.79kg)}$

$w_2=0.28Hz$

$f_2=2\pi w_2$

$f_2=1.76Hz$

$T_2=1/f_2$

$T_2=1/1.76Hz$

$T_2=0.58s$   :)