Question

In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.517 ss for the value of the period.
Trial 1 Spring constant is 117N/m, period of oscillations .37s, mass of the block is .400kg .
Trial 2 oscillation period is .52s

Answer 1

Answer:

$M_2=0.79kg$

Explanation:

From the question we are told that:

Period $T=0.517s$

Trial 1

Spring constant $\mu=117N/m$

Period $T_1=0.37$

Mass $m=0.400kg$

Trial 2

Period $T_2=0.52$

Generally the equation for Spring Constant  is mathematically given by

\mu=\frac{4 \pi^2 M}{T^2}

Since

$\mu _1=\mu_2$

Therefore

$\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}$

$M_2=M_1*(\frac{T_2}{T_1})^2$

$M_2=0.400*(\frac{0.52}{0.37}})^2$

$M_2=0.79kg$