Edit to “Solution”
en/7.4.35.md
+1 −1
| ### Statement | |||
| $7.4.35.$ What is the period of small vibrations of four charged bodies connected by | |||
| identical filaments of length l and moving as shown in the figure? Mass and | |||
| charge of the body m and q. | |||
|  | |||
| ### Solution | |||
| The First Way | |||
| Force Way | |||
| The diagonal | |||
| $$ | |||
| d=l\times\sqrt(2) | |||
| $$ | |||
| $$ | |||
| d1=d-x | |||
| $$ | |||
| @@ -23,7 +23,7 @@Solution | |||
| There are 2 ways to find out difference of force | |||
| first | |||
| $$ | |||
| − | \DeltaF=F2-F1=k×q×q×(1/(d×d)-1(d1×d1)) | ||
| + | \Delta F=F2-F1=k×q×q×(1/(d×d)-1(d1×d1)) | ||
| $$ | |||
| The second way | |||
| $$ | |||
| F=k×q×q/(d1×d1) | |||
| $$ | |||
| $$ | |||
| dF/dx=-2k×q×q/(d-x)\wedge3 | |||
| $$ | |||
| There is picture for dF | |||
|  | |||
| k' is stifness of spring | |||
| $$ | |||
| k'×dx=dF | |||
| $$ | |||
| $$ | |||
| k'=2×k×q×q×dx/d\wedge3 | |||
| $$ | |||
| But the spring located between 2 object.We need period one of them | |||
| The period formula | |||
| $$ | |||
| T=2×\Pi×\sqrt(m/K) | |||
| $$ | |||
| But the spring is between 2 object | |||
| We need use half of spring | |||
| so | |||
| $$ | |||
| K=k'×2 | |||
| $$ | |||
| The period is | |||
| $$ | |||
| T=2×\Pi×\sqrt(m×d\wedge3/(4×k×q×q)) | |||
| $$ | |||
|  | |||
| #### Answer | |||
| $$ | |||
| T=2×\Pi×\sqrt(4×\Pi×\varepsilon×m×l×l/\sqrt(2)×q×q) | |||
| $$ | |||
| unchanged lines 38 | |||
| ### Statement | ### Statement | ||
| $7.4.35.$ What is the period of small vibrations of four charged bodies connected by | $7.4.35.$ What is the period of small vibrations of four charged bodies connected by | ||
| identical filaments of length l and moving as shown in the figure? Mass and | identical filaments of length l and moving as shown in the figure? Mass and | ||
| charge of the body m and q. | charge of the body m and q. | ||
|  |  | ||
| ### Solution | ### Solution | ||
| The First Way | The First Way | ||
| Force Way | Force Way | ||
| The diagonal | The diagonal | ||
| $$ | $$ | ||
| d=l\times\sqrt(2) | d=l\times\sqrt(2) | ||
| $$ | $$ | ||
| $$ | $$ | ||
| d1=d-x | d1=d-x | ||
| $$ | $$ | ||
| @@ -23,7 +23,7 @@Solution | |||
| There are 2 ways to find out difference of force | There are 2 ways to find out difference of force | ||
| first | first | ||
| $$ | $$ | ||
| \DeltaF=F2-F1=k×q×q×(1/(d×d)-1(d1×d1)) | \Delta F=F2-F1=k×q×q×(1/(d×d)-1(d1×d1)) | ||
| $$ | $$ | ||
| The second way | The second way | ||
| $$ | $$ | ||
| F=k×q×q/(d1×d1) | F=k×q×q/(d1×d1) | ||
| $$ | $$ | ||
| $$ | $$ | ||
| dF/dx=-2k×q×q/(d-x)\wedge3 | dF/dx=-2k×q×q/(d-x)\wedge3 | ||
| $$ | $$ | ||
| There is picture for dF | There is picture for dF | ||
|  |  | ||
| k' is stifness of spring | k' is stifness of spring | ||
| $$ | $$ | ||
| k'×dx=dF | k'×dx=dF | ||
| $$ | $$ | ||
| $$ | $$ | ||
| k'=2×k×q×q×dx/d\wedge3 | k'=2×k×q×q×dx/d\wedge3 | ||
| $$ | $$ | ||
| But the spring located between 2 object.We need period one of them | But the spring located between 2 object.We need period one of them | ||
| The period formula | The period formula | ||
| $$ | $$ | ||
| T=2×\Pi×\sqrt(m/K) | T=2×\Pi×\sqrt(m/K) | ||
| $$ | $$ | ||
| But the spring is between 2 object | But the spring is between 2 object | ||
| We need use half of spring | We need use half of spring | ||
| so | so | ||
| $$ | $$ | ||
| K=k'×2 | K=k'×2 | ||
| $$ | $$ | ||
| The period is | The period is | ||
| $$ | $$ | ||
| T=2×\Pi×\sqrt(m×d\wedge3/(4×k×q×q)) | T=2×\Pi×\sqrt(m×d\wedge3/(4×k×q×q)) | ||
| $$ | $$ | ||
|  |  | ||
| #### Answer | #### Answer | ||
| $$ | $$ | ||
| T=2×\Pi×\sqrt(4×\Pi×\varepsilon×m×l×l/\sqrt(2)×q×q) | T=2×\Pi×\sqrt(4×\Pi×\varepsilon×m×l×l/\sqrt(2)×q×q) | ||
| $$ | $$ | ||
| unchanged lines 38 | |||