| ### Solution | | ### Solution |
| The initial velocity is zero, therefore | | The initial velocity is zero, therefore |
| $$ | | $$ |
| v=\int_0^t a \, dt \tag{1} | | v=\int_0^t a \, dt \tag{1} |
| $$ | | $$ |
| Newton's second law: | | Newton's second law: |
| $$ | | $$ |
| F=-Ee=am_e \tag{2} | | F=-Ee=am_e \tag{2} |
| $$ | | $$ |
| Let $a_0=\frac{eE_0}{m_e}$, then from (2) into (1): | | Let $a_0=\frac{eE_0}{m_e}$, then from (2) into (1): |
| $$ | | $$ |
| v(t)=-a_0\int_0^t \sin (\omega t + \varphi) \, dt=\frac{eE_0}{m_e\omega} \left(\cos (\omega t + \varphi)-\cos \varphi\right) \tag{3} | | v(t)=-a_0\int_0^t \sin (\omega t + \varphi) \, dt=\frac{eE_0}{m_e\omega} \left(\cos (\omega t + \varphi)-\cos \varphi\right) \tag{3} |
| $$ | | $$ |
| Taking into account the range of the cosine, the expression in parentheses can take values $[-1-\cos \varphi;\, 1-\cos \varphi]$. One must be careful here: remember that we need the maximum magnitude of the velocity, and that $\cos \varphi$ can be either positive or negative. Finally, analysing (3), we obtain | | Taking into account the range of the cosine, the expression in parentheses can take values $[-1-\cos \varphi;\, 1-\cos \varphi]$. One must be careful here: remember that we need the maximum magnitude of the velocity, and that $\cos \varphi$ can be either positive or negative. Finally, analysing (3), we obtain |
| $$ | | $$ |
| v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|). | | v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|). |
| $$ | | $$ |
| It is clear that the velocity changes periodically with period $T=\frac{2\pi}{\omega}$. Then the average value is | | It is clear that the velocity changes periodically with period $T=\frac{2\pi}{\omega}$. Then the average value is |
| $$ | | $$ |
| \bar{v}=\frac{1}{T}\int_0^T v(t) \, dt \tag{4} | | \bar{v}=\frac{1}{T}\int_0^T v(t) \, dt \tag{4} |
| $$ | | $$ |
| I will allow myself to skip the calculation of the integral – remembering that the integral of cosine over its period is zero, this can be done very quickly. Eventually, taking the modulus, we get: | | I will allow myself to skip the calculation of the integral – remembering that the integral of cosine over its period is zero, this can be done very quickly. Eventually, taking the modulus, we get: |
| $$ | | $$ |
| \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi. | | \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi. |
| $$ | | $$ |
| | | |
| There is a typo in the author's answer. | | There is a typo in the author's answer. |
| | | |
| #### Answer | | #### Answer |
| $$ | | $$ |
| v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|) | | v_{max}=\frac{eE_0}{m_e\omega}(1+|\cos\varphi|) |
| $$ | | $$ |
| $$ | | $$ |
| \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi | | \bar{v}=\frac{2eE_0}{m_e\omega}\cos\varphi |
| $$ | | $$ |
