物原2复习提纲

振动


一、振动方程:\(y=Acos(\omega t+\varphi)\)

(\(\omega =2\pi \nu=\frac{2\pi}{T}\) )


二、波动方程:\(y(x,t)=Acos(2\pi\nu t+\varphi-\frac{2\pi(x-x_0)}{\lambda})\)=\(Acos(2\pi\nu t+\varphi-\frac{\omega(x-x_0)}{\mu})\)

(波向x正方向传播则\(\varphi\)后是负号,否则是正号);


三、反射波:
假设入射波为:\(y(x,t)=Acos(2\pi\nu t+\varphi-\frac{2\pi x}{\lambda})\),则

\(y'(x,t)=\left\{\begin{array}
固定端反射:y(x,t)=Acos(2\pi\nu t+\varphi+\frac{2\pi x}{\lambda}+\pi),有半波损失 \\
自由端反射:y(x,t)=Acos(2\pi\nu t+\varphi+\frac{2\pi x}{\lambda}),无半波损失
\end{array} \right.\)


四、驻波:\(y=y_1+y_2=2Acos(\frac{2\pi x}{\lambda}+\frac{\varphi_2-\varphi_1}{2})cos(2\pi\nu t+\frac{\varphi_2+\varphi_1}{2})\)

(注意式中的\(\varphi_2-\varphi_1\)是向x轴负方向的波-向x轴正方向的波)

波节:\(cos(\frac{2\pi x}{\lambda}+\frac{\varphi_2-\varphi_1}{2})=0\)处,动能为0;

波腹:\(cos(\frac{2\pi x}{\lambda}+\frac{\varphi_2-\varphi_1}{2})=1\)处,振动势能为0;


五、光的干涉:

干涉条件:
1)频率相同 2)振动方向平行 3)相位差恒定

光程差:

相位差:\(\Delta\varphi=\frac{2\pi r_2}{\lambda_2}-\frac{2\pi r_1}{\lambda_1}+\varphi_1-\varphi_2\)
又有\(\lambda_1=\frac{\lambda}{n_1},\lambda_2=\frac{\lambda}{n_2}\)
所以\(\Delta\varphi=2\pi(\frac{n_2 r_2}{\lambda}-\frac{n_1 r_1}{\lambda})\)
则光程差\(\epsilon=n_2 r_2-n_1 r_1)\)
光程差与相位差的关系是\(\Delta\varphi=\frac{2\pi}{\lambda} \epsilon\)

\(\Delta\varphi=\pm 2k\pi\) 时,干涉相长

\(\Delta\varphi=\pm (2k+1)\pi\) 时,干涉相消

杨氏干涉:

明条纹中心位置:\(x=\pm k\lambda\frac{D}{d}\); \(\epsilon=\pm k\lambda\)

暗条纹中心位置:\(x=\pm (2k+1)\frac{\lambda}{2}\frac{D}{d}\); \(\epsilon=\pm (2k+1)\frac{\lambda}{2}\)

劈尖干涉:

干涉条纹位于上表面;

明纹条件:\(\epsilon=2n_2 h+\frac{\lambda}{2}=k\lambda\)
暗纹条件:\(\epsilon=2n_2 h+\frac{\lambda}{2}=(2k+1)\lambda\)

\(h_k -h_{k+1}=\frac{\lambda}{2n_2}\)

条纹间距:\(a=\frac{\lambda}{2n_2 sin\theta}\)

牛顿环干涉:

内疏外密

明环条件:\(n\frac{r^2}{R}+\frac{\lambda}{2}=k\lambda, r=\sqrt{(2k-1)\frac{R\lambda}{2n}}\)

暗环条件:\(n\frac{r^2}{R}+\frac{\lambda}{2}=(2k+1)\frac{\lambda}{2}, r=\sqrt{k\frac{R\lambda}{n}}\)


单缝衍射:

明纹条件:\(a sin\varphi=\pm (2k+1)\frac{\lambda}{2}\)
暗纹条件:\(a sin\varphi=\pm 2k\frac{\lambda}{2}\)
明纹范围:\(-\frac{\lambda}{a}<sin\varphi<\frac{\lambda}{a}\)

中央亮纹宽度:\(\Delta x_0=2f\frac{\lambda}{a}\)
其他亮纹宽度:\(\Delta x=f\frac{\lambda}{a}\)

光栅衍射:

亮纹处合振幅为N,合光强为\(N^2\)

亮纹条件:\(dsin\varphi=(a+b)sin\varphi=\pm k\lambda\)

缺级条件:\(asin\varphi=\pm 2k’\frac{\lambda}{2}\)
所以\(k=\frac{a+b}{a}k’\)时缺级

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