112 年 國立中山大學材料與光電科學學系碩士班《普通物理學》

📄 試題原卷 免費註冊後即可對照原始考卷 PDF免費註冊

第 1. 題40 分

Problem 1. [Mechanics: 40 points]
In the following, we consider particles only move in x-direction.
(a) [5 points] There are two springs, one with the elastic constant k1k_1, and the other with elastic constant k2k_2. If the two springs are in series, what is the effective elastic constant keffk_{eff}?
(b) [5 points] Consider 2 identical atoms with mass mm, connected by a chemical bond modeled as a spring with elastic constant k0k_0 (a sigma bond). The center of mass is stationary (not moving). Solve the angular frequency of the vibration motion of the 2 atoms. (hint: it might be easier if using the result of (a))
(c) [5 points] There are two springs, one with the elastic constant k1k_1, and the other with elastic constant k2k_2. If the two springs are in parallel, what is the effective elastic constant keffk_{eff}?
(d) [5 points] Consider 2 identical atoms with mass mm, connected by 2 chemical bonds (that is, a double bond) modeled as 2 springs with elastic constant k0k_0. The center of mass is stationary (not moving). Solve the angular frequency of the vibration motion of the 2 atoms. (hint: use the result of (c))
(e) [10 points] Consider 2 non-identical atoms with mass MM and mm (such as a carbon and an oxygen atom), connected by a chemical bond modeled as a spring with elastic constant k0k_0 (a sigma bond). The center of mass is stationary (not moving). Solve the angular frequency of the vibration motion of the 2 atoms.
(f) [5 points] Consider 2 non-identical atoms with mass MM and mm (such as a carbon and an oxygen atom), connected by 3 chemical bonds modeled as 3 springs with the same elastic constant k0k_0 (a sigma bond). (i) solve the angular frequency of the vibration motion of the 2 atoms, and (ii) compare the obtained frequency with the single bond case above.
(g) [5 points] Carbon dioxide CO2CO_2 can be modeled as one carbon atom and two oxygen atoms linked by springs (double bonds). How many are the vibrational modes? Please give a proper argument, not just giving the answer.

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

本題核心是「等效彈簧常數」與「簡諧運動」:

  • 串聯彈簧承受相同的力,總伸長量為各彈簧伸長量之和。
  • 並聯彈簧的伸長量相同,總恢復力為各彈簧恢復力之和。
  • 雙原子振動可用相對座標描述,其等效質量為約化質量
    μ=MmM+m\mu=\frac{Mm}{M+m}
  • 振動角頻率為
    ω=keffμ\omega=\sqrt{\frac{k_{\mathrm{eff}}}{\mu}}

題目限定粒子只沿 xx 方向運動,因此只需討論縱向振動。


解題方法

(a)兩彈簧串聯

串聯時兩彈簧承受相同的力 FF。設兩彈簧伸長量分別為 x1x_1、x2x_2,總伸長量為 xx:

x=x1+x2x=x_1+x_2

由虎克定律:

F=k1x1=k2x2F=k_1x_1=k_2x_2

因此:

x1=Fk1,x2=Fk2x_1=\frac{F}{k_1},\qquad x_2=\frac{F}{k_2}

總伸長量為:

x=F(1k1+1k2)x=F\left(\frac{1}{k_1}+\frac{1}{k_2}\right)

定義等效彈簧常數 keffk_{\mathrm{eff}}:

F=keffxF=k_{\mathrm{eff}}x

所以:

1keff=1k1+1k2\frac{1}{k_{\mathrm{eff}}} = \frac{1}{k_1}+\frac{1}{k_2}

得到:

keff=k1k2k1+k2\boxed{k_{\mathrm{eff}}=\frac{k_1k_2}{k_1+k_2}}

(b)兩個質量相同的原子,以一個化學鍵相連

兩原子質量皆為 mm,化學鍵的彈簧常數為 k0k_0。

質心固定,因此兩原子的位移大小相等、方向相反。令兩原子的位移為:

x1=u,x2=−ux_1=u,\qquad x_2=-u

化學鍵的伸長量為兩端相對位移:

Δx=x1−x2=2u\Delta x=x_1-x_2=2u

第一個原子受到的恢復力為:

F=−k0(x1−x2)=−2k0uF=-k_0(x_1-x_2)=-2k_0u

由牛頓第二定律:

mu¨=−2k0um\ddot u=-2k_0u

整理為簡諧運動方程:

u¨+2k0mu=0\ddot u+\frac{2k_0}{m}u=0

因此角頻率為:

ω1=2k0m\boxed{\omega_1=\sqrt{\frac{2k_0}{m}}}

也可使用約化質量:

μ=m⋅mm+m=m2\mu=\frac{m\cdot m}{m+m}=\frac{m}{2}

由 ω=k0/μ\omega=\sqrt{k_0/\mu} 同樣得到:

ω1=k0m/2=2k0m\omega_1=\sqrt{\frac{k_0}{m/2}} =\sqrt{\frac{2k_0}{m}}

(c)兩彈簧並聯

並聯時兩彈簧的伸長量相同,設共同伸長量為 xx。

各彈簧的恢復力為:

F1=k1x,F2=k2xF_1=k_1x,\qquad F_2=k_2x

總恢復力:

F=F1+F2=(k1+k2)xF=F_1+F_2=(k_1+k_2)x

因此等效彈簧常數為:

keff=k1+k2\boxed{k_{\mathrm{eff}}=k_1+k_2}

(d)兩個相同原子,以雙鍵相連

雙鍵由兩條相同的化學鍵構成,每條彈簧常數均為 k0k_0。兩條彈簧並聯,因此:

keff=k0+k0=2k0k_{\mathrm{eff}}=k_0+k_0=2k_0

兩原子的約化質量仍為:

μ=m2\mu=\frac{m}{2}

故角頻率:

ω2=keffμ=2k0m/2\omega_2 = \sqrt{\frac{k_{\mathrm{eff}}}{\mu}} = \sqrt{\frac{2k_0}{m/2}}

因此:

ω2=2k0m\boxed{\omega_2=2\sqrt{\frac{k_0}{m}}}

與單鍵結果比較:

ω2ω1=2k0/m2k0/m=2\frac{\omega_2}{\omega_1} = \frac{2\sqrt{k_0/m}}{\sqrt{2k_0/m}} = \sqrt{2}

所以雙鍵的振動角頻率是單鍵的 2\sqrt{2} 倍。


(e)兩個不同質量的原子,以單鍵相連

兩原子質量分別為 MM 與 mm,單鍵彈簧常數為 k0k_0。

系統的約化質量為:

μ=MmM+m\mu=\frac{Mm}{M+m}

振動角頻率為:

ω=k0μ\omega = \sqrt{\frac{k_0}{\mu}}

代入約化質量:

ω=k0(M+m)Mm\omega = \sqrt{\frac{k_0(M+m)}{Mm}}

因此:

ωsingle=k0(M+m)Mm\boxed{\omega_{\text{single}} = \sqrt{\frac{k_0(M+m)}{Mm}}}

也可由質心固定直接推導。設兩原子位移為 xMx_M 與 xmx_m,質心固定代表:

MxM+mxm=0Mx_M+mx_m=0

兩原子的相對位移為:

q=xM−xmq=x_M-x_m

系統的相對運動方程為:

μq¨=−k0q\mu\ddot q=-k_0q

因此:

q¨+k0μq=0\ddot q+\frac{k_0}{\mu}q=0

所得角頻率即為上式。


(f)兩個不同質量的原子,以三重鍵相連

🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 2. 題35 分

Problem 2. [Electromagnetism: 35 points]
The electric field at a distance r from a point charge Q is:
E=14πϵ0Qr2r^E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \hat{r}
where r^\hat{r} is the unit vector along the vector r⃗\vec{r}. The electric potential is:
V=14πϵ0QrV = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}
Consider the following questions in the vacuum, and use the units in the above formula.
(a) [5 points] If there are two point charges −q-q and qq, separated by a distance dd, as shown in the figure below. The magnitude of the dipole moment p⃗\vec{p} of these two point charges is then p=qdp = qd. Write down the electric potential at rr (the origin is at the middle of the two charges), using those quantities given in the figure.
(b) [5 points] Prove V=14πϵ0pcos⁡θr2V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}.
(c) [5 points] Use the approximation (1+δ)α≈1+αδ(1+\delta)^\alpha \approx 1+\alpha\delta if δ≪1\delta \ll 1 to show: when r≫dr \gg d (measuring the potential far away from the dipole), V≈14πϵ0qdr2(1±12dr)V \approx \frac{1}{4\pi\epsilon_0} \frac{qd}{r^2} (1 \pm \frac{1}{2}\frac{d}{r}).
(d) [10 points] When r≫dr \gg d (measuring the potential far away from the dipole), show the electric potential V=14πϵ0pcos⁡θr2V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}.
(e) [10 points] From the above result, calculate the electric field far away from the dipole using:
E⃗=−∇V=−(∂V∂rr^+1r∂V∂θθ^+1rsin⁡θ∂V∂ϕϕ^)\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial r}\hat{r} + \frac{1}{r}\frac{\partial V}{\partial\theta}\hat{\theta} + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\hat{\phi} \right)

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

本題考查電偶極子的:

  • 電位疊加原理;
  • 電偶極矩定義;
  • 二項式近似
    (1+δ)α≈1+αδ,∣δ∣≪1(1+\delta)^\alpha \approx 1+\alpha\delta,\qquad |\delta|\ll 1
  • 球座標中的梯度,以及由電位求電場:
    E⃗=−∇V\vec E=-\nabla V

令偶極子中心為原點,偶極軸取為 zz 軸,+q+q 位於 z=+d/2z=+d/2,−q-q 位於 z=−d/2z=-d/2。偶極矩方向由 −q-q 指向 +q+q,因此

p⃗=qd z^,p=qd\vec p=qd\,\hat z,\qquad p=qd

觀測點與偶極軸的夾角為 θ\theta。


(a) 電位的精確表示式

由電位的疊加原理,兩個點電荷產生的總電位為

V=V+q+V−qV=V_{+q}+V_{-q}

觀測點到 +q+q 與 −q-q 的距離分別為

r+=r2+d24−rdcos⁡θr_+=\sqrt{r^2+\frac{d^2}{4}-rd\cos\theta}

以及

r−=r2+d24+rdcos⁡θr_-=\sqrt{r^2+\frac{d^2}{4}+rd\cos\theta}

因此

V=14πϵ0(qr+−qr−)V=\frac{1}{4\pi\epsilon_0} \left( \frac{q}{r_+}-\frac{q}{r_-} \right)

即

V=q4πϵ0[1r2+d24−rdcos⁡θ−1r2+d24+rdcos⁡θ]\boxed{ V= \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{\sqrt{r^2+\frac{d^2}{4}-rd\cos\theta}} - \frac{1}{\sqrt{r^2+\frac{d^2}{4}+rd\cos\theta}} \right] }

這是任意位置的精確電位。


(b) 證明遠處電位為 V=14πϵ0pcos⁡θr2V=\dfrac{1}{4\pi\epsilon_0}\dfrac{p\cos\theta}{r^2}

當觀測距離遠大於偶極子長度時,r≫dr\gg d,可將兩個距離近似為

r+≈r−d2cos⁡θr_+\approx r-\frac{d}{2}\cos\theta r−≈r+d2cos⁡θr_-\approx r+\frac{d}{2}\cos\theta

因此

V≈q4πϵ0[1r−d2cos⁡θ−1r+d2cos⁡θ]V\approx \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r-\frac{d}{2}\cos\theta} - \frac{1}{r+\frac{d}{2}\cos\theta} \right]

令

δ=dcos⁡θ2r\delta=\frac{d\cos\theta}{2r}

則

1r−d2cos⁡θ=1r11−δ≈1r(1+δ)\frac{1}{r-\frac{d}{2}\cos\theta} = \frac{1}{r}\frac{1}{1-\delta} \approx \frac{1}{r}(1+\delta)

而

1r+d2cos⁡θ=1r11+δ≈1r(1−δ)\frac{1}{r+\frac{d}{2}\cos\theta} = \frac{1}{r}\frac{1}{1+\delta} \approx \frac{1}{r}(1-\delta)

相減得

V≈q4πϵ01r[(1+δ)−(1−δ)]V\approx \frac{q}{4\pi\epsilon_0} \frac{1}{r} \left[(1+\delta)-(1-\delta)\right] V≈q4πϵ02δrV\approx \frac{q}{4\pi\epsilon_0} \frac{2\delta}{r}

代入 δ=dcos⁡θ/(2r)\delta=d\cos\theta/(2r):

V≈14πϵ0qdcos⁡θr2V\approx \frac{1}{4\pi\epsilon_0} \frac{qd\cos\theta}{r^2}

由 p=qdp=qd,得到

V≈14πϵ0pcos⁡θr2\boxed{ V\approx \frac{1}{4\pi\epsilon_0} \frac{p\cos\theta}{r^2} }

這就是電偶極子的遠場電位。


(c) 使用二項式近似

精確距離可寫成

r+=r1−drcos⁡θ+d24r2r_+=r\sqrt{1-\frac{d}{r}\cos\theta+\frac{d^2}{4r^2}} r−=r1+drcos⁡θ+d24r2r_-=r\sqrt{1+\frac{d}{r}\cos\theta+\frac{d^2}{4r^2}}

當 r≫dr\gg d 時,保留一階小量:

r+≈r1−drcos⁡θr_+\approx r\sqrt{1-\frac{d}{r}\cos\theta} r−≈r1+drcos⁡θr_-\approx r\sqrt{1+\frac{d}{r}\cos\theta}

利用

(1+δ)−1/2≈1−δ2(1+\delta)^{-1/2}\approx 1-\frac{\delta}{2}

可得

1r+≈1r(1+12drcos⁡θ)\frac{1}{r_+} \approx \frac{1}{r} \left( 1+\frac{1}{2}\frac{d}{r}\cos\theta \right)

以及

1r−≈1r(1−12drcos⁡θ)\frac{1}{r_-} \approx \frac{1}{r} \left( 1-\frac{1}{2}\frac{d}{r}\cos\theta \right)

因此,兩個電荷各自的電位近似為

V+q≈14πϵ0qr(1+12drcos⁡θ)V_{+q}\approx \frac{1}{4\pi\epsilon_0} \frac{q}{r} \left( 1+\frac{1}{2}\frac{d}{r}\cos\theta \right)
🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 3. 題10 分

Problem 3. [Thermodynamics: 10 points]
(a) [5 points] A gas is enclosed in a cylinder with a moveable piston, and follows the equation: p3V5=constantp^3V^5 = \text{constant}.
where PP is the pressure, and VV is the volume of the gas. Find the work done when the volume is
compressed from VAV_A to VBV_B, with a known initial pressure PAP_A (the final pressure PBP_B is not known).
(b) [5 points] The Maxwell-Boltzmann distribution of the speed v of the particle is:
f(v)=(m2πkBT)3/24πv2e−mv22kBTf(v) = \left(\frac{m}{2\pi k_B T}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2k_B T}}
where mm is the mass of the particle, kBk_B is the Boltzmann constant, and TT is the temperature. Calculate
the most probable speed vpv_p, which corresponds to the maximum value of f(v)f(v).

登入後即可作答並保存紀錄。

這一題的完整詳解

本題主要考察熱力學中的功的計算以及統計物理中的速度分佈函數求極值。

3. (a) 氣體壓縮過程的功
此題描述了一個氣體遵循 p3V5=constantp^3V^5 = \text{constant} 的過程,並要求計算從 VAV_A 壓縮到 VBV_B 所做的功。
設 p3V5=Cp^3V^5 = C,其中 CC 是常數。
因此,p=C1/3V−5/3p = C^{1/3} V^{-5/3}。
功的定義為 W=∫VAVBp dVW = \int_{V_A}^{V_B} p \, dV。
將 pp 的表達式代入:
W=∫VAVBC1/3V−5/3 dVW = \int_{V_A}^{V_B} C^{1/3} V^{-5/3} \, dV
W=C1/3∫VAVBV−5/3 dVW = C^{1/3} \int_{V_A}^{V_B} V^{-5/3} \, dV
計算積分:
∫V−5/3 dV=V−5/3+1−5/3+1=V−2/3−2/3=−32V−2/3\int V^{-5/3} \, dV = \frac{V^{-5/3 + 1}}{-5/3 + 1} = \frac{V^{-2/3}}{-2/3} = -\frac{3}{2} V^{-2/3}
所以:
W=C1/3[−32V−2/3]VAVBW = C^{1/3} \left[ -\frac{3}{2} V^{-2/3} \right]_{V_A}^{V_B}
W=−32C1/3(VB−2/3−VA−2/3)W = -\frac{3}{2} C^{1/3} (V_B^{-2/3} - V_A^{-2/3})
我們需要將常數 CC 用已知的初始條件 PAP_A 和 VAV_A 來表示。
由 p3V5=Cp^3V^5 = C,我們有 C=PA3VA5C = P_A^3 V_A^5。
代入 C1/3=(PA3VA5)1/3=PAVA5/3C^{1/3} = (P_A^3 V_A^5)^{1/3} = P_A V_A^{5/3}。
W=−32(PAVA5/3)(VB−2/3−VA−2/3)W = -\frac{3}{2} (P_A V_A^{5/3}) (V_B^{-2/3} - V_A^{-2/3})
W=−32PAVA5/3VB−2/3+32PAVA5/3VA−2/3W = -\frac{3}{2} P_A V_A^{5/3} V_B^{-2/3} + \frac{3}{2} P_A V_A^{5/3} V_A^{-2/3}
W=−32PA(VAVB)2/3VA+32PAVAW = -\frac{3}{2} P_A \left(\frac{V_A}{V_B}\right)^{2/3} V_A + \frac{3}{2} P_A V_A
W=32PAVA−32PA(VAVB)2/3VAW = \frac{3}{2} P_A V_A - \frac{3}{2} P_A \left(\frac{V_A}{V_B}\right)^{2/3} V_A
W=32PAVA[1−(VAVB)2/3]W = \frac{3}{2} P_A V_A \left[ 1 - \left(\frac{V_A}{V_B}\right)^{2/3} \right]
另外一種表達方式:
由於 p3V5=Cp^3V^5 = C,所以 pV5/3=C1/3p V^{5/3} = C^{1/3}。
PAVA5/3=PBVB5/3P_A V_A^{5/3} = P_B V_B^{5/3}。
PB=PA(VAVB)5/3P_B = P_A \left(\frac{V_A}{V_B}\right)^{5/3}。
將 C1/3C^{1/3} 替換成 PBVB5/3P_B V_B^{5/3}。
W=−32PBVB5/3(VB−2/3−VA−2/3)W = -\frac{3}{2} P_B V_B^{5/3} (V_B^{-2/3} - V_A^{-2/3})
W=−32PBVB5/3VB−2/3+32PBVB5/3VA−2/3W = -\frac{3}{2} P_B V_B^{5/3} V_B^{-2/3} + \frac{3}{2} P_B V_B^{5/3} V_A^{-2/3}
W=−32PBVB+32PBVB5/3VA−2/3W = -\frac{3}{2} P_B V_B + \frac{3}{2} P_B V_B^{5/3} V_A^{-2/3}
W=−32PBVB+32PA(VAVB)5/3VB5/3VA−2/3=−32PBVB+32PA(VAVB)5/3VB5/3VA−2/3W = -\frac{3}{2} P_B V_B + \frac{3}{2} P_A \left(\frac{V_A}{V_B}\right)^{5/3} V_B^{5/3} V_A^{-2/3} = -\frac{3}{2} P_B V_B + \frac{3}{2} P_A \left(\frac{V_A}{V_B}\right)^{5/3} V_B^{5/3} V_A^{-2/3}

🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 4. 題15 分

Problem 4. [Waves/Optics/Modern physics: 15 points]
If a propagating wave along the x direction has the form f(x,t)=Aei(kx−ωt+ϕ)f(x,t) = A e^{i(kx-\omega t+\phi)}, where xx is the
position, tt is the time, AA is a constant amplitude and ϕ\phi is a constant phase.
(a) [5 points] (i) Show that this wave function satisfies the wave equation
∂2f∂x2=1v2∂2f∂t2\frac{\partial^2 f}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 f}{\partial t^2}
(ii) express the velocity vv using AA, kk, ω\omega, or ϕ\phi.
(b) [5 points] The wave is a periodic function of the time, so f(x,t0)=f(x,t0+T)f(x, t_0) = f(x, t_0+T) for arbitrary t0t_0
(where TT is the period). Use this relation, express the period TT with AA, kk, ω\omega, and ϕ\phi.
(c) [5 points] The wave is a periodic function of the position, so f(x0,t)=f(x0+λ,t)f(x_0, t) = f(x_0+\lambda, t) for arbitrary
x0x_0 (where λ\lambda is the wave length). Use this relation, express the wave length λ\lambda with AA, kk, ω\omega, and ϕ\phi.

登入後即可作答並保存紀錄。

這一題的完整詳解

本題主要考察平面波的波動方程、週期和波長的計算。

4. (a) 波動函數的波動方程驗證與速度求解
給定的波函數為 f(x,t)=Aei(kx−ωt+ϕ)f(x,t) = A e^{i(kx-\omega t+\phi)}。
(i) 驗證波動方程 ∂2f∂x2=1v2∂2f∂t2\frac{\partial^2 f}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 f}{\partial t^2}。
首先計算對 xx 的二階偏導數:
∂f∂x=A⋅ikei(kx−ωt+ϕ)\frac{\partial f}{\partial x} = A \cdot i k e^{i(kx-\omega t+\phi)}
∂2f∂x2=A⋅(ik)2ei(kx−ωt+ϕ)=A⋅(−k2)ei(kx−ωt+ϕ)=−k2f(x,t)\frac{\partial^2 f}{\partial x^2} = A \cdot (ik)^2 e^{i(kx-\omega t+\phi)} = A \cdot (-k^2) e^{i(kx-\omega t+\phi)} = -k^2 f(x,t)
接下來計算對 tt 的二階偏導數:
∂f∂t=A⋅(−iω)ei(kx−ωt+ϕ)\frac{\partial f}{\partial t} = A \cdot (-i\omega) e^{i(kx-\omega t+\phi)}
∂2f∂t2=A⋅(−iω)2ei(kx−ωt+ϕ)=A⋅(−ω2)ei(kx−ωt+ϕ)=−ω2f(x,t)\frac{\partial^2 f}{\partial t^2} = A \cdot (-i\omega)^2 e^{i(kx-\omega t+\phi)} = A \cdot (-\omega^2) e^{i(kx-\omega t+\phi)} = -\omega^2 f(x,t)
將這兩個結果代入波動方程:
−k2f(x,t)=1v2(−ω2f(x,t))-k^2 f(x,t) = \frac{1}{v^2} (-\omega^2 f(x,t))
−k2=−ω2v2-k^2 = -\frac{\omega^2}{v^2}
k2=ω2v2k^2 = \frac{\omega^2}{v^2}
v2=ω2k2v^2 = \frac{\omega^2}{k^2}
v=±ωkv = \pm \frac{\omega}{k}
由於波沿 xx 方向傳播,速度 vv 可以是正的或負的。通常我們取正值表示沿 xx 正方向傳播,即 v=ωkv = \frac{\omega}{k}。

(ii) 表達速度 vv。
從上面的推導,我們得到 v2=ω2k2v^2 = \frac{\omega^2}{k^2}。
因此,速度 vv 為:
v=ωkv = \frac{\omega}{k}
注意,速度 vv 的值與振幅 AA 和相位 ϕ\phi 無關。

🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

其他考古題