Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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]

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.