Engineering Physics I | Semester 1 BTech CSE Core Notes

Semester 1 - Page 2: Engineering Physics I. General first-year concept and exam-practice notes for mechanics, properties of matter, oscillations, waves and thermal physics. Course placement varies. These are not an official syllabus or predictions of exam questions. Optics, electromagnetism, quantum physics and electronic materials continue in Physics II; consult your own outline to decide the semester.

Study method: define the system, draw the diagram, list assumptions, select the formula, substitute SI units, then check the result. A conservation law needs the right system and conditions.

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Topic map

1. Measurement, units, dimensions and errors
2. Vectors, resolution and motion
3. Newton laws, free-body diagrams and friction
4. Projectile motion and circular motion
5. Work, energy, potential and power
6. Impulse, momentum, centre of mass and collisions
7. Rotation, moment of inertia, torque and angular momentum
8. Rolling, equilibrium and gravitational motion
9. Elasticity and properties of matter
10. Simple harmonic motion: equation, phase and energy
11. Pendulums and small-angle approximation
12. Damping, forced oscillation and resonance
13. Travelling waves, standing waves and energy
14. Sound, decibels, Doppler effect and ultrasonics
15. Temperature, heat, expansion and calorimetry
16. Ideal gases, kinetic theory and heat capacities
17. First law, work and thermodynamic processes
18. Second law, entropy, heat engines and heat transfer

Draw forces acting on the chosen block. The action-reaction partners belong on the other bodies, not this diagram.

1. Measurement, units, dimensions and errors

Use SI base units consistently: metre, kilogram, second and kelvin appear most often here. Dimensional consistency is necessary, but cannot determine a dimensionless constant such as 2 or pi. Report measured answers with sensible significant figures. An absolute uncertainty has the same unit as the quantity; fractional uncertainty is the ratio of uncertainty to the quantity. For independent random uncertainties added in quadrature, the assumptions differ from a worst-case error bound. State which method you use.

Worked example: A length (2.00 +/- 0.01) m has fractional uncertainty 0.005, or 0.5%. For a worst-case first-order estimate of area L squared, relative error is approximately twice that in L: 1%. The dimensional formula of force is M L T^(-2), not M L T^(-1).

Exam Q: Can dimensions alone prove a formula correct?
Answer: No. They catch unit inconsistency but not missing dimensionless factors or wrong physical assumptions.

2. Vectors, resolution and motion

A scalar has magnitude; a vector also has direction. Resolve vectors into perpendicular components with signs before combining them. Dot products give a scalar; cross products give a vector perpendicular to the operands, with the right-hand rule. Displacement differs from total distance. Velocity is the time derivative of position, and acceleration the derivative of velocity. The equations v=u+at, s=ut+at squared/2 and v squared=u squared+2as require constant acceleration.

Worked example: A force of 10 N at 30 degrees above horizontal has components 8.660 N and 5 N. A body starting at 2 m/s with constant acceleration 3 m/s squared reaches 14 m/s after 4 s and moves 32 m. Returning to the starting point gives zero displacement but can involve nonzero distance.

Exam Q: Why can acceleration exist at constant speed?
Answer: Velocity includes direction. Uniform circular motion changes direction continuously, so it has centripetal acceleration.

3. Newton laws, free-body diagrams and friction

In an inertial frame an object keeps constant velocity when net external force is zero. For constant mass, sum of external forces equals ma. An action-reaction pair acts on different bodies, so it does not cancel on a single-body free-body diagram. Draw the chosen body alone, name every external force, choose axes, and write a separate component equation for each axis. Static friction adjusts up to mu_s N; it is not always equal to that limit. Kinetic friction is often modelled as mu_k N.

Worked example: For a 2 kg block on a horizontal surface, take g=9.8 m/s squared. N=19.6 N if there is no other vertical force. With a 10 N horizontal pull and mu_k=0.2, sliding friction is 3.92 N and acceleration is (10-3.92)/2=3.04 m/s squared. If it is not yet sliding, first check the static-friction limit.

Exam Q: Is the normal reaction always mg?
Answer: No. Inclines, vertical components of applied forces and accelerating supports change the normal force. Find it from the relevant force equation.

4. Projectile motion and circular motion

For ideal projectile motion, ignore air resistance and use uniform gravitational acceleration. Horizontal and vertical motions share time but have different accelerations. The familiar range and time-of-flight formulas assume launch and landing at the same height. For uniform circular motion, acceleration v squared/r points toward the centre; centripetal force is the net inward force, not an extra force to add to tension or gravity. A centrifugal term appears in a rotating-frame description, not as an extra physical force in an inertial-frame FBD.

Worked example: For u=20 m/s at 30 degrees, g=9.8 m/s squared and equal launch/landing height: time=2.041 s, range=35.348 m, maximum height=5.102 m. For mass 0.5 kg moving at 4 m/s on radius 2 m, net inward force=0.5(16)/2=4 N.

Exam Q: When does R=u squared sin(2theta)/g fail?
Answer: When landing height differs, air resistance matters, or the constant-g model is unsuitable. Use component equations with the actual endpoint conditions.

5. Work, energy, potential and power

For a constant force, work is F dot displacement=Fd cos theta. For varying force, use the path integral. Net work equals change in kinetic energy, K=mv squared/2. For conservative forces, work equals the negative change in potential energy: near Earth, U=mgh; for an ideal spring, U=kx squared/2. Mechanical energy stays constant only when no nonconservative work changes it. Total energy can remain conserved even as friction reduces mechanical energy. Instantaneous power is F dot v; average power is work/time.

Worked example: A 10 N force at 60 degrees to a displacement of 2 m does 10 J. A 2 kg body speeding from 2 to 4 m/s gains 12 J of kinetic energy. A spring with k=200 N/m compressed 0.1 m stores 1 J. A 100 W motor delivering useful work for 5 s supplies 500 J.

Exam Q: Does friction destroy energy?
Answer: No. It transfers mechanical energy into internal energy and other forms. Do not confuse mechanical-energy conservation with total-energy conservation.

6. Impulse, momentum, centre of mass and collisions

Momentum is p=mv for a nonrelativistic particle. External impulse is the time integral of net external force and equals the change in total momentum of the chosen system. Momentum is conserved when net external impulse on that system is zero or negligible over the interval. Internal collision forces cancel in the system momentum balance. An elastic collision also conserves kinetic energy; an inelastic collision does not. Centre-of-mass position is the mass-weighted average, and its motion obeys M a_CM=F_external.

Worked example: A 2 kg trolley at 3 m/s sticks to a 1 kg trolley at rest. Neglect horizontal external impulse: final speed=(2x3+1x0)/3=2 m/s. Kinetic energy changes from 9 J to 6 J, a reduction of 3 J. This becomes other energy forms, including deformation, heat and sound. Masses 2 kg at x=0 and 1 kg at x=3 m have centre of mass x=1 m.

Exam Q: Is momentum conserved in every collision for any chosen system?
Answer: No. You must account for external impulse and choose the system boundary. One trolley alone exchanges momentum with the other; an isolated pair conserves total momentum.

7. Rotation, moment of inertia, torque and angular momentum

Angular displacement is in radians. For a fixed axis, v=r omega and tangential acceleration=r alpha. Moment of inertia is sum or integral of r perpendicular squared times mass and depends on the axis. For a rigid body about a fixed axis, torque=I alpha and rotational kinetic energy=I omega squared/2. Angular momentum about a specified origin is conserved when net external torque about that origin is zero. The parallel-axis theorem I=I_CM+Md squared uses parallel axes. The perpendicular-axis theorem is restricted to a planar lamina.

Worked example: A solid disc of mass 2 kg and radius 0.5 m has central-axis I=MR squared/2=0.25 kg m squared. At omega=4 rad/s its rotational energy is 2 J. A torque of 1 N m gives alpha=4 rad/s squared. Shifting to a parallel tangent axis adds MR squared=0.5, giving I=0.75 kg m squared.

Exam Q: Why must an axis be named with moment of inertia?
Answer: The distances in I are measured from that axis. The same body generally has different moments of inertia about different axes.

8. Rolling, equilibrium and gravitational motion

Pure rolling on a stationary surface requires v_CM=R omega and no slipping. Its kinetic energy has translation and rotation parts. Static equilibrium requires both net force and net torque to vanish. Choose a torque point that removes unknown reaction forces when possible. Universal gravitation has magnitude Gm1m2/r squared. For a circular orbit around a dominant mass M, balancing gravity with centripetal force gives v=sqrt(GM/r). Kepler laws describe ideal two-body orbital behaviour; orbital period squared is proportional to semimajor axis cubed for a fixed central mass.

Worked example: A solid disc rolling at centre speed 2 m/s, with M=2 kg and R=0.5 m, has omega=4 rad/s. Translational energy is 4 J and rotational energy 2 J: total 6 J. A horizontal beam supported at both ends with a centred 100 N load, neglecting its own weight, has reactions 50 N at each end.

Exam Q: Can force balance alone establish static equilibrium?
Answer: No. A nonzero net couple can rotate the body even when the resultant force is zero. Check torque balance too.

9. Elasticity and properties of matter

Stress is force divided by area; normal strain is fractional length change. In the linear elastic region, Young modulus E=normal stress/normal strain. Shear and bulk moduli relate shear stress to shear strain and pressure change to fractional volume compression. Hooke law is a small-deformation model, not a rule beyond the elastic limit. Poisson ratio is minus transverse strain divided by axial strain. For an isotropic linear elastic material, E=2G(1+nu)=3K(1-2nu). These relations need the material-model assumptions.

Worked example: A 1000 N axial force on area 1x10^(-4) m squared produces stress 10 MPa. With E=200 GPa, strain is 5x10^(-5). A 2 m bar extends 0.0001 m, or 0.1 mm. Doubling length at fixed force, area and E doubles extension; it does not change the material modulus.

Exam Q: What is the difference between stiffness and Young modulus?
Answer: Young modulus describes a material under the model. A component stiffness also depends on geometry, such as EA/L for an axially loaded uniform bar.

10. Simple harmonic motion: equation, phase and energy

SHM has acceleration proportional to displacement and directed toward equilibrium: a=-omega squared x. For an ideal spring, m x double-prime=-kx, so omega=sqrt(k/m), T=2pi sqrt(m/k), and x=A cos(omega t+phi). Velocity is -A omega sin(omega t+phi); acceleration is -omega squared x. A sets the amplitude and phi the initial phase. Total ideal-oscillator energy is kA squared/2; it moves between kinetic and spring energy. Do not confuse omega in rad/s with frequency f in Hz: omega=2pi f.

Worked example: With m=1 kg, k=100 N/m and A=0.1 m: omega=10 rad/s, T=0.6283 s, f=1.5915 Hz, maximum speed=1 m/s, and total energy=0.5 J. At x=0.06 m, potential energy=0.18 J, kinetic=0.32 J and speed=0.8 m/s. At either extreme speed is zero, not acceleration.

Exam Q: Derive the spring period from Newton law.
Answer: Insert the restoring force -kx into m a. Then a=-(k/m)x, identify omega squared=k/m, and use T=2pi/omega.

11. Pendulums and small-angle approximation

A simple pendulum is a point mass on a light, inextensible string. Its exact angular restoring term contains sin theta; replacing sin theta by theta requires small angles in radians. Then T=2pi sqrt(L/g), approximately independent of mass and amplitude. A physical pendulum uses T=2pi sqrt(I/(mgd)), where d is pivot-to-centre-of-mass distance. A torsional pendulum has torque=-kappa theta and T=2pi sqrt(I/kappa). Use the moment of inertia about the actual pivot.

Worked example: For a simple pendulum of length 1 m with g=9.8 m/s squared, the small-angle period is 2.007 s. Quadrupling length doubles period. A torsional oscillator with I=0.02 kg m squared and kappa=0.5 N m/rad has period 1.257 s. A larger simple-pendulum amplitude requires a correction beyond the small-angle result.

Exam Q: Why is mass absent from the ideal simple-pendulum period?
Answer: Both inertia and gravitational restoring torque are proportional to the bob mass, so it cancels in the small-angle equation.

12. Damping, forced oscillation and resonance

A common model is m x double-prime+b x prime+kx=F0 cos(Omega t). Free underdamped motion has an exponentially decreasing amplitude, with decay factor exp(-bt/(2m)) and frequency sqrt(k/m-(b/(2m)) squared). Critical damping returns to equilibrium without oscillation as fast as this linear model allows; stronger damping is overdamped. A driving force produces a steady-state response at the driving frequency after transients decay. Resonance describes a large response near a system frequency; damping makes it finite. The displacement-amplitude peak is slightly below the undamped natural frequency when damping is weak enough for a peak to exist.

Worked example: For m=1 kg, k=100 N/m and b=2 kg/s, the free damped angular frequency is sqrt(99)=9.950 rad/s. The amplitude after 1 s is exp(-1)=0.3679 times its initial value. More damping dissipates energy faster per given velocity, but excessive damping makes the return slow.

Exam Q: Why should bridges and machines avoid troublesome resonance?
Answer: Repeated driving can build a large vibration response, stressing components. Change stiffness/mass, reduce the forcing, or add damping; the suitable remedy depends on the system.

13. Travelling waves, standing waves and energy

A sinusoidal wave may be written y=A sin(kx-omega t+phi) for travel in the positive x direction. Here k=2pi/lambda is wave number, unlike the spring constant used earlier. Wave speed v=omega/k=f lambda. A taut-string wave has speed sqrt(tension/linear mass density) in the ideal model. Standing waves form by superposition of oppositely travelling waves and have stationary nodes and antinodes. A string fixed at both ends has allowed frequencies n v/(2L). Waves transport energy; the medium particles need not travel with the wave.

Worked example: For tension 100 N and linear density 0.01 kg/m, string wave speed is 100 m/s. A 1 m fixed-end string has fundamental frequency 50 Hz and second harmonic 100 Hz. A wave of frequency 20 Hz and wavelength 5 m travels at 100 m/s; its period is 0.05 s.

Exam Q: Does every periodic particle motion constitute a travelling wave?
Answer: No. A travelling wave requires a disturbance propagating through space. A single oscillator can move periodically without creating such propagation.

14. Sound, decibels, Doppler effect and ultrasonics

Sound in a fluid is a longitudinal pressure wave. Intensity is power per area and sound intensity level is beta=10 log10(I/I0), conventionally I0=10^(-12) W/m squared. Equal independent incoherent sounds add intensities, not their decibel numbers. For a stationary observer and a source approaching in still air, f_observed=f_source v_sound/(v_sound-v_source), using positive approaching source speed below the sound speed. Ultrasonics are sound above human hearing, conventionally above about 20 kHz; applications include imaging, cleaning and defect detection. Echo distances use round-trip time.

Worked example: If intensity is 10^(-6) W/m squared, level is 60 dB. Doubling that intensity gives approximately 63.01 dB, not 120 dB. An ultrasound echo returning after 0.0002 s in material with sound speed 5000 m/s indicates a reflector 0.5 m away: distance=vt/2.

Exam Q: Why is an echo distance divided by two?
Answer: The measured time includes travel to the reflector and back. The one-way distance is half the total travelled distance.

15. Temperature, heat, expansion and calorimetry

Temperature describes thermal state; heat names energy transfer caused by a temperature difference. Internal energy belongs to the system; heat is not a stored substance. Zeroth law gives the transitive basis of thermal equilibrium and temperature measurement. For approximately constant specific heat without phase change, Q=mc delta T. A phase change uses Q=mL under the stated conditions and can occur without a temperature increase. Small linear expansion is delta L=alpha L0 delta T. Kelvin and Celsius differences are equal in size, but gas-law temperatures must be absolute kelvin.

Worked example: Heating 0.5 kg water by 10 K with c=4180 J/(kg K) takes 20900 J before losses. A 2 m aluminium rod with alpha=23x10^(-6)/K warmed by 50 K lengthens 2.3 mm. Mixing equal masses of water at 20 and 60 degrees Celsius gives 40 degrees Celsius if heat loss and container heat capacity are negligible.

Exam Q: Why can supplied heat fail to raise temperature?
Answer: During a phase change it changes the internal state and energy rather than increasing temperature, under the relevant equilibrium conditions.

16. Ideal gases, kinetic theory and heat capacities

The ideal-gas equation is pV=nRT with kelvin temperature. It is a model that works best when intermolecular effects are small. For a classical monatomic ideal gas, internal energy U=3nRT/2, molar Cv=3R/2 and Cp=5R/2. More molecular degrees of freedom can contribute; do not apply these values to every gas or every temperature. The root-mean-square molecular speed is sqrt(3RT/M), with molar mass M in kg/mol. For an ideal gas Cp-Cv=R per mole and gamma=Cp/Cv.

Worked example: One mole of ideal gas at 300 K in volume 0.024 m cubed has pressure about 103925 Pa using R=8.314 J/(mol K). Its monatomic-model internal energy is 3741.3 J. Warming one mole of this model by 10 K at constant volume requires 124.71 J.

Exam Q: Why must molar mass be in kg/mol for the SI rms-speed formula?
Answer: R uses SI energy units. Inserting grams without conversion gives inconsistent units and a speed wrong by a factor of sqrt(1000).

17. First law, work and thermodynamic processes

For a closed system with the usual thermal-work model, delta U=Q-W, taking Q positive into the system and W positive done by the system. Some books use work-on-system and a plus sign; state the convention. Quasi-static expansion work is integral p dV. Constant-volume work is zero; constant-pressure work is p delta V. An isothermal ideal-gas process has delta U=0, so Q=W. An adiabatic process has Q=0, not necessarily constant temperature. For a reversible adiabatic ideal gas with constant heat capacities, pV^gamma is constant.

Worked example: A gas absorbs 500 J and does 200 J work: delta U=300 J. For reversible isothermal expansion of one mole at 300 K to twice its volume, W=nRT ln2=1728.85 J and Q has the same value. At constant volume, adding 100 J increases internal energy by 100 J if no other energy transfer occurs.

Exam Q: Is adiabatic the same as isothermal?
Answer: No. Adiabatic means no heat transfer. Isothermal means constant temperature. An adiabatically expanding ideal gas normally cools as it does work.

18. Second law, entropy, heat engines and heat transfer

An engine cycle has zero net internal-energy change. Efficiency is net work/heat absorbed=1-Q_cold/Q_hot for magnitudes. A reversible engine between reservoirs at T_hot and T_cold has Carnot efficiency 1-T_cold/T_hot, with kelvin temperatures; no engine between those reservoirs exceeds it. Entropy is a state function. Its change can be found along a reversible path by integral delta Q_reversible/T. An isolated system total entropy cannot decrease; entropy of a nonisolated subsystem can decrease while the surroundings compensate. Conduction, convection and radiation are different heat-transfer routes.

Worked example: For hot and cold reservoirs at 600 K and 300 K, Carnot efficiency is 0.5. An engine absorbing 1000 J per cycle cannot deliver more than 500 J under those reservoir conditions. Adding 100 J reversibly at fixed 300 K changes entropy by 1/3 J/K. A reversible ideal-gas cycle has zero net system entropy change, not zero heat flow at every step.

Exam Q: Why does the first law alone not decide engine efficiency?
Answer: It balances energy but does not set the direction and limits of thermal processes. The second law supplies the entropy and efficiency constraints.

Difference table: answer with conditions

Pair	First	Second	Common mistake
Elastic / inelastic collision	Kinetic energy conserved	Kinetic energy changes	Both need negligible external impulse for total momentum conservation
Heat / temperature	Energy transfer; joule	Thermal state; kelvin	Saying heat is a substance stored in a body
Isothermal / adiabatic	Temperature constant	Heat transfer zero	Assuming both mean delta U=0
Travelling / standing wave	Pattern propagates	Nodes stay fixed	Confusing wave speed with particle speed
Force / torque	Changes momentum; N	Changes angular momentum; N m	Using force balance but forgetting moment balance

Practice set with solved answers

1. A 3 kg object has 12 N net force. Acceleration=4 m/s squared.
2. A spring mass doubles at fixed k. Period increases by sqrt(2).
3. A 0.1 kg ball changes velocity from -10 to +10 m/s. Impulse=+2 N s.
4. An ideal-gas cycle takes in 800 J and rejects 500 J. Net work=300 J; efficiency=37.5%.
5. A reversible adiabatic process has Q=0. Does its temperature stay fixed? No; expansion usually lowers it for an ideal gas.
6. A fixed-end string fundamental is 40 Hz. Third harmonic=120 Hz.

Exam checklist: name the frame and system; distinguish net force from applied force; state collision impulse conditions; use radians and kelvin where needed; name rotation axes; show the heat/work sign convention; include units and a sensible number of significant figures. A correct number without conditions can answer a different problem.

Coverage and reference reading

This page develops the earlier mechanics, SHM and heat summary into 18 worked topics. It includes a bridge to gravity, elasticity, acoustics and thermodynamics commonly placed in first-year outlines. It does not replace full derivations, textbook exercises or a college-specific syllabus. Additional advanced treatments such as Lagrangian mechanics, coupled normal modes and relativistic dynamics depend on your course.

Read the textbook explanations for momentum conservation, rotation, elasticity, SHM, waves, first law and engines. Examples and answered questions here are original study practice, not copied past papers.

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