Oscillations Numerical Solutions

phy-1109 PHY-1109 Physics

I. SHM Superposition

1. Two SHM Superposition (2016)

Problem: $Y_1=2\sin (\omega t+\pi /4)$ and $Y_2=3\sin (\omega t+\pi /3)$.

Find: (i) Amplitude, (ii) Phase constant, (iii) Time period.

Solution:

We have two waves with amplitudes $A_1=2$, $A_2=3$ and phase angles ${\varphi }_1=45^{\circ }(\pi /4)$, ${\varphi }_2=60^{\circ }(\pi /3)$.

The phase difference $\delta ={\varphi }_2-{\varphi }_1=60^{\circ }-45^{\circ }=15^{\circ }$.

(i) Resultant Amplitude ($R$):

R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos(\delta)}$$$$R = \sqrt{2^2 + 3^2 + 2(2)(3)\cos(15^\circ)}$$$$R = \sqrt{4 + 9 + 12(0.9659)}$$$$R = \sqrt{13 + 11.59} = \sqrt{24.59} \approx \mathbf{4.96}

(ii) Phase Constant ($\Theta$):

\tan\Theta = \frac{A_1\sin\phi_1 + A_2\sin\phi_2}{A_1\cos\phi_1 + A_2\cos\phi_2}$$$$\tan\Theta = \frac{2\sin(45^\circ) + 3\sin(60^\circ)}{2\cos(45^\circ) + 3\cos(60^\circ)}$$$$\tan\Theta = \frac{2(0.707) + 3(0.866)}{2(0.707) + 3(0.5)} = \frac{1.414 + 2.598}{1.414 + 1.5} = \frac{4.012}{2.914} \approx 1.376$$$$\Theta = \tan^{-1}(1.376) \approx 54^\circ \text{ or } \mathbf{0.3\pi \text{ rad}}

(iii) Time Period ($T$):

$T=\frac{2\pi }{\omega }$

(Answer depends on the value of $\omega$, which is symbolic here.)

2. Two SHM Superposition (2024)

Problem: $y_1=5\sin (\omega t+\pi /2)$, $y_2=6\sin (\omega t+\pi /3)$.

Find: (i) Amplitude, (ii) Phase constant, (iii) Time period.

Solution:

$A_1=5,{\varphi }_1=90^{\circ }$. $A_2=6,{\varphi }_2=60^{\circ }$.

Phase difference $\delta =90^{\circ }-60^{\circ }=30^{\circ }$.

(i) Resultant Amplitude ($R$):

R = \sqrt{5^2 + 6^2 + 2(5)(6)\cos(30^\circ)}$$$$R = \sqrt{25 + 36 + 60(0.866)} = \sqrt{61 + 51.96} = \sqrt{112.96} \approx \mathbf{10.63}

(ii) Phase Constant ($\Theta$):

\tan\Theta = \frac{5\sin(90^\circ) + 6\sin(60^\circ)}{5\cos(90^\circ) + 6\cos(60^\circ)}$$$$\tan\Theta = \frac{5(1) + 6(0.866)}{5(0) + 6(0.5)} = \frac{5 + 5.196}{3} = \frac{10.196}{3} \approx 3.398$$$$\Theta = \tan^{-1}(3.398) \approx 73.6^\circ \text{ or } \mathbf{0.41\pi \text{ rad}}

(iii) Time Period: $T=2\pi /\omega$.

II. SHM Calculation

3. SHM Parameters (2018)

Problem: $y=12\sin (6t-\pi /3)$.

Find: Frequency, epoch, max displacement, velocity & acceleration at $t=0,6$.

Solution:

Comparing to $y=A\sin (\omega t+\varphi )$:

• Max Displacement ($A$): 12 units

• Angular Frequency ($\omega$): 6 rad/s

• Frequency ($f$): $\omega /2\pi =6/2\pi \approx 0.955\text{ Hz}$

• Epoch (Initial Phase): $-\pi /3$ rad or $-60^{\circ }$

Velocity ($v$) and Acceleration ($a$):

v = \frac{dy}{dt} = 12(6)\cos(6t - \pi/3) = 72\cos(6t - \pi/3)$$$$a = \frac{d^2y}{dt^2} = -72(6)\sin(6t - \pi/3) = -432\sin(6t - \pi/3)

• At $t=0$:

  • $v(0)=72\cos (-\pi /3)=72(0.5)=36$

  • $a(0)=-432\sin (-\pi /3)=-432(-\sqrt{3}/2)=216\sqrt{3}\approx 374.12$

• At $t=6$:

  • Angle $\theta =36-\pi /3$ rad. ($36\text{ rad}\approx 2062.6^{\circ }$, $\pi /3\approx 60^{\circ }$). ${\theta }_{net}\approx 2002.6^{\circ }$.

  • $\text{Mod }360^{\circ }\approx 202.6^{\circ }$.

  • $v(6)=72\cos (36-\pi /3)\approx -66.5$

  • $a(6)=-432\sin (36-\pi /3)\approx 166.4$

4. SHM at Specific Position (2021)

Problem: $A=15$ cm, $f=4$ Hz. Find max $v,a$ and speed at $x=9$ cm.

Solution:

$\omega =2\pi f=8\pi$ rad/s.

• Max Velocity: $v_{max}=\omega A=8\pi (15)=120\pi \approx 376.99\text{ cm/s}$

• Max Acceleration: $a_{max}={\omega }^{2}A=(8\pi {)}^2(15)=64{\pi }^2(15)=960{\pi }^2\approx 9474.8{\text{ cm/s}}^2$

• Speed at $x=9$:

  v = \omega\sqrt{A^2 - x^2}$$$$v = 8\pi\sqrt{15^2 - 9^2} = 8\pi\sqrt{225 - 81} = 8\pi\sqrt{144} = 8\pi(12)$$$$v = 96\pi \approx \mathbf{301.59 \text{ cm/s}}

III. Work Done (Forced Oscillation)

General Formula:

Work done over time $t$ for sustained forced oscillation is $W=P_{avg}\times t$.

Average Power $P_{avg}=\frac{F_0{v}_0}{2}\cos (\varphi )$, where $\varphi$ is the phase difference between Force and Velocity.

• $y=A\sin (\omega t-\theta )$

• $v=A\omega \cos (\omega t-\theta )=A\omega \sin (\omega t-\theta +\pi /2)$.

• Phase of Force = 0. Phase of Velocity = $-\theta +\pi /2$.

• Phase diff $\varphi =\theta -\pi /2$.

• Alternatively, $P_{avg}=\frac{F_0\omega A}{2}\sin (\theta )$ where $\theta$ is the phase lag of displacement with respect to force.

5. Work Done (2019)

Problem: $F=5\sin (\pi t)$, $y=0.50\sin (\pi t-\pi /4)$. $t_1=30s,t_2=5m$.

Solution:

$F_0=5,A=0.5,\omega =\pi$. Phase lag $\theta =\pi /4$.

$P_{avg}=\frac{5\times \pi \times 0.5}{2}\sin (\pi /4)=\frac{2.5\pi }{2}(0.707)\approx 1.25\pi (0.707)\approx 2.77\text{ Watts}$

• (i) $W_{30s}=2.77\times 30=83.1\text{ Joules}$

• (ii) $W_{5min}=2.77\times (5\times 60)=2.77\times 300=831\text{ Joules}$

6. Work Done (2020)

Problem: $F=25\sin (\pi t/10)$, $Y=0.5\sin (\pi t/10-\pi /4)$. $t_1=30s,t_2=5m$.

Solution:

$F_0=25,A=0.5,\omega =\pi /10$. Phase lag $\theta =\pi /4$.

$P_{avg}=\frac{25\times (\pi /10)\times 0.5}{2}\sin (\pi /4)=\frac{1.25\pi }{2}(0.707)\approx 0.625\pi (0.707)\approx 1.388\text{ Watts}$

• (i) $W_{30s}=1.388\times 30\approx 41.6\text{ Joules}$

• (ii) $W_{5min}=1.388\times 300\approx 416.4\text{ Joules}$

7. Work Done (2022)

Problem: $F=0.5\sin (2t)$, $Y=0.5\sin (2t-\pi /2)$. $t_1=12m,t_2=1\text{ day}$.

Solution:

$F_0=0.5,A=0.5,\omega =2$. Phase lag $\theta =\pi /2$ (Resonance condition).

$P_{avg}=\frac{0.5\times 2\times 0.5}{2}\sin (\pi /2)=\frac{0.5}{2}(1)=0.25\text{ Watts}$

• (i) $W_{12min}=0.25\times (12\times 60)=0.25\times 720=180\text{ Joules}$

• (ii) $W_{1day}=0.25\times (24\times 3600)=0.25\times 86400=21,600\text{ Joules}$

IV. Waves

8. Wave Parameters (2018)

Problem: $y=0.02\sin (30t-4x)$.

Solution:

Compare to $y=A\sin (\omega t-kx)$.

• Amplitude: $0.02\text{ units}$

• $\omega =30⟹f=30/2\pi \approx 4.77\text{ Hz}$

• $k=4⟹\lambda =2\pi /4=1.57\text{ units}$

• Speed: $v=\omega /k=30/4=7.5\text{ units/sec}$

9. Stationary Waves (2019)

Problem: $y=0.8\sin (0.32x)\cos (16t)$.

Solution:

Standard form: $y=2A\sin (kx)\cos (\omega t)$.

• Resultant Amplitude term: $0.8$.

• Amplitude of individual waves ($A$): $0.8/2=0.4$

• $\omega =16,k=0.32$.

• Velocity ($v$): $\omega /k=16/0.32=50\text{ units/sec}$

V. Damped Oscillation & Resonance

10. Damped Oscillation (2017, 2020)

Problem: $A$ reduced to $1/10$th after 100 oscillations. $T=1.25s$.

Solution:

Formula: $A_n=A_0{e}^{-\gamma t}$, where $\gamma$ is the decay constant.

Time $t=100\times T=100\times 1.25=125$ seconds.

$A(125)=0.1A_0$.

0.1 = e^{-125\gamma} \implies \ln(0.1) = -125\gamma$$$$-2.302 = -125\gamma \implies \gamma = 0.0184 \text{ s}^{-1}

• (i) Damping Constant ($\gamma$): $0.0184{\text{ s}}^{-1}$ (Note: if defined as $k$ in $e^{-kt}$, answer is the same. If defined as $b$ where $\gamma =b/2m$, mass is needed).

• (i) Relaxation Time ($\tau$): Defined for energy decay $\tau =1/2\gamma$.

  $\tau =1/(2\times 0.0184)\approx 27.17\text{ sec}$

• (ii) Amplitude without damping: This simply refers to the initial amplitude $A_0$ (which is unknown but constant). If the question meant “Natural Period”, $T_0\approx T_d$ for small damping.

11 & 12. LCR Resonance (2019, 2024)

Formulas:

Resonant Frequency $f_0=\frac{1}{2\pi \sqrt{LC}}$

Quality Factor $Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

Q11 (2019): $L=2mH,C=50\mu F,R=2\Omega$

• $f_0=\frac{1}{2\pi \sqrt{2\times 10^{-3}\times 50\times 10^{-6}}}=\frac{1}{2\pi \sqrt{100\times 10^{-9}}}=\frac{1}{2\pi (3.16\times 10^{-4})}\approx 503\text{ Hz}$

• $Q=\frac{1}{2}\sqrt{\frac{2\times 10^{-3}}{50\times 10^{-6}}}=0.5\sqrt{40}=0.5(6.32)\approx 3.16$

Q12 (2024): $L=5.2mH,C=5\mu F,R=0.35\Omega$

• $f_0=\frac{1}{2\pi \sqrt{5.2\times 10^{-3}\times 5\times 10^{-6}}}=\frac{1}{2\pi \sqrt{26\times 10^{-9}}}\approx 987\text{ Hz}$

• $Q=\frac{1}{0.35}\sqrt{\frac{5.2\times 10^{-3}}{5\times 10^{-6}}}=2.857\sqrt{1040}=2.857(32.25)\approx 92.1$

13. Forced Oscillation (2021)

Problem: $m=2g,b=8$ dyne-s/cm. Force $F=250\sin (2t)$.

Solution:

Damping factor $\gamma =b/2m=8/(2\times 2)=2$.

• Velocity Resonance: Occurs at natural frequency ${\omega }_0$.

• Displacement Resonance: Occurs at ${\omega }_r=\sqrt{{\omega }_0^2-2{\gamma }^2}$.

• Note: The problem does not provide the spring constant $k$ or natural frequency ${\omega }_0$. Numerical calculation is impossible without it. If $k$ was given, plug into formulas above.

14. Damping Check (2023)

Problem: $C=0.1\mu F,L=10mH,R=210\Omega$.

Solution:

Condition for oscillation: $R^2<\frac{4L}{C}$.

• $R^2=210^2=44,100$

• $4L/C=\frac{4\times 10\times 10^{-3}}{0.1\times 10^{-6}}=\frac{40\times 10^{-3}}{10^{-7}}=400,000$

• Since $44,100<400,000$, the discharge is oscillatory.

• Frequency: $f=\frac{1}{2\pi }\sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}\approx 5033\text{ Hz}$

VI. Acoustics & Doppler

16. Doppler’s Effect (2016)

Problem: Two planes pass each other.

Status: Incomplete data (missing velocities/frequencies).

Formula: $f^{\prime }=f\left(\frac{v\pm v_{observer}}{v\mp v_{source}}\right)$.

17. Doppler’s Effect (2022)

Problem: Ratio of apparent frequencies 6:5. Engine passes stationary observer.

Solution:

Before passing (Approach): $f_1=f\frac{v}{v-v_s}$

After passing (Recede): $f_2=f\frac{v}{v+v_s}$

Ratio $f_1/f_2=\frac{v+v_s}{v-v_s}=\frac{6}{5}$.

5(v+v_s) = 6(v-v_s) \implies 5v + 5v_s = 6v - 6v_s \implies v = 11v_s$$$$v_s = v/11

Assuming speed of sound $v=330$ m/s: $v_s=30$ m/s.

18 & 19. Intensity Levels

Formula: $\Delta L=10{\log }_{10}(I_2/I_1)$

• Q18 (2017): Sound doubled ($I_2=2I_1$). $\Delta L=10\log (2)\approx 3.01\text{ dB}$

• Q19 (2023): Sound $\times 100$ ($I_2=100I_1$). $\Delta L=10\log (100)=20\text{ dB}$

20 & 21. Acoustic Intensity Calculations

Problem: Intensity Level (IL) = 80 dB (Q20) / 100 dB (Q21). Ref $I_0=10^{-12}$.

Formulas:

1. $IL=10\log (I/I_0)⟹I=I_0\cdot 10^{(IL/10)}$

2. Acoustic Pressure $P_{rms}=\sqrt{I\cdot \rho v}$. (Assume $\rho v\approx 415$ Rayls for air).

Q20 (80 dB):

• $I=10^{-12}\cdot 10^8=10^{-4}{\text{ W/m}}^2$

• $P_{rms}=\sqrt{10^{-4}\times 415}\approx 0.2\text{ Pa}$

Q21 (100 dB):

• $I=10^{-12}\cdot 10^{10}=10^{-2}{\text{ W/m}}^2$

• $P_{rms}=\sqrt{10^{-2}\times 415}\approx 2.04\text{ Pa}$

22 & 23. Sound Energy Flow

Problem: Calculate energy flow per unit area ($I$).

Formula: $I=2{\pi }^2{f}^2{A}^2\rho v$. (Assume $\rho =1.29{\text{ kg/m}}^3,v=330\text{ m/s}$).

Q22 (2024 becm): $f=250,A=0.4\text{ cm}=0.004\text{ m}$.

I = 2\pi^2 (250)^2 (0.004)^2 (1.29)(330)$$$$I \approx 19.74 \times 62500 \times 0.000016 \times 425.7 \approx \mathbf{8405 \text{ W/m}^2}

To find per $c{m}^2$: Divide by $10^4\to 0.84{\text{ W/cm}}^2$.

Q23 (2024): $f=256,A=5\times 10^{-5}\text{ m}$.

I = 2\pi^2 (256)^2 (5\times 10^{-5})^2 (425.7)$$$$I \approx 19.74 \times 65536 \times (25 \times 10^{-10}) \times 425.7 \approx \mathbf{0.00137 \text{ W/m}^2}

25. Reverberation (2023)

Problem: Room $6\times 5\times 4$ m. $V=120{\text{ m}}^3$. Surface $S=2(30+24+20)=148{\text{ m}}^2$.

• (i) Mean free path ($\lambda$):

  $\lambda =\frac{4V}{S}=\frac{4\times 120}{148}\approx 3.24\text{ meters}$

• (ii) Reflections per second ($n$):

  $n=\frac{v}{\lambda }=\frac{330}{3.24}\approx 101.8\text{ reflections/sec}$