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astra-53 [7]
3 years ago
13

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 0.51

7 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
Physics
1 answer:
Masteriza [31]3 years ago
5 0

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

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