Edit to “Solution”

astrosander edited
revision #12565 parent #11814 ← older
@@ -6,11 +6,11 @@Statement
### Solution
−The period of oscillation of a spring pendulum is found as $$ T=2\pi\sqrt{\frac{m}{k}}\quad(1) $$ Next, we will find and use the equivalent stiffness for each spring system under study
+The period of oscillation of a spring pendulum is found as $$ T=2\pi\sqrt{\frac{m}{k}}\tag{1} $$ Next, we will find and use the equivalent stiffness for each spring system under study
a) Based on the results from [2.1.15](../../2/2.1.15), we found that when springs are connected in parallel, their equivalent stiffness $$ k'=k_1+k_2 $$ Substituting into $(1)$: $$ T_1=2\pi\sqrt{\frac{m}{k_1+k_2}} $$ b) Alternatively, from [2.1.16](../2.1.16), we obtained that when springs are connected in parallel, their equivalent stiffness is $$ k'=\frac{k_1\cdot k_2}{k_1+k_2} $$ Substituting into $(1)$: $$ T_2=2\pi\sqrt{\frac{m(k_1+k_2)}{k_1\cdot k_2}} $$ c) Also in [2.1.16](../2.1.16), we showed that this scheme is equivalent to the case of parallel connection of springs
−![ Part of the solution from $2.1.16$ |1153x419, 67%](../../img/3.2.4/3.2.4_1.png)
+![ For solution $2.1.16$ |1153x419, 60%](../../img/3.2.4/3.2.4_1.png)
Equivalent spring stiffness $$ k'=k_1+k_2 $$ Substituting into $(1)$: $$ T_3=2\pi\sqrt{\frac{m}{k_1+k_2}} $$
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