Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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



Sagot :

Answer:

[tex]M_2=0.79kg[/tex]

Explanation:

From the question we are told that:

Period [tex]T=0.517s[/tex]

Trial 1

Spring constant [tex]\mu=117N/m[/tex]

Period [tex]T_1=0.37[/tex]

Mass [tex]m=0.400kg[/tex]

Trial 2

Period [tex]T_2=0.52[/tex]

Generally the equation for Spring Constant  is mathematically given by

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

Since

[tex]\mu _1=\mu_2[/tex]

Therefore

[tex]\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}[/tex]

[tex]M_2=M_1*(\frac{T_2}{T_1})^2[/tex]

[tex]M_2=0.400*(\frac{0.52}{0.37}})^2[/tex]

[tex]M_2=0.79kg[/tex]