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GRE Physics Subject Test study materials
GRE Physics Subject Test study resources

GRE Physics Subject Test practice questions

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Question 1Classical Mechanics

A block slides down a frictionless incline of angle θ. What is its acceleration along the incline?

  1. g sin θ Correct answer.
  2. g cos θ
  3. g tan θ
  4. g

Explanation

On a frictionless incline the only force component along the surface is the tangential part of gravity, mg sin θ, so dividing by mass gives a = g sin θ. The normal force balances the perpendicular component mg cos θ and contributes nothing along the incline.

Question 2Classical Mechanics

A projectile is launched over level ground and lands at the same height. Neglecting air resistance, which quantity remains constant throughout the flight?

  1. The horizontal component of velocity Correct answer.
  2. The speed of the projectile
  3. The kinetic energy of the projectile
  4. The vertical component of velocity

Explanation

With no air resistance there is no horizontal force, so the horizontal velocity component is constant. Gravity acts vertically, changing the vertical component, the speed, and the kinetic energy continuously.

Question 3Classical Mechanics

A wheel spins from rest with constant angular acceleration 2 rad/s². Through what angle does it turn in 3 s?

  1. 18 rad
  2. 12 rad
  3. 6 rad
  4. 9 rad Correct answer.

Explanation

For constant angular acceleration starting from rest, θ = ½αt² = ½(2)(3)² = 9 rad. The linear analog s = ½at² applies directly with angular quantities.

Question 4Classical Mechanics

Two isolated particles attract each other gravitationally. Which statement about the forces they exert on each other is correct?

  1. The force on each particle is proportional to its own mass
  2. The more massive particle pulls harder on the less massive one
  3. The forces are equal in magnitude and opposite in direction Correct answer.
  4. The forces are equal in magnitude and in the same direction

Explanation

Newton's third law requires the mutual gravitational forces to be equal in magnitude and opposite in direction, regardless of the masses. Each force has magnitude Gm₁m₂/r², which depends on both masses, not on either one alone.

Question 5Classical Mechanics

A 2 kg object moves at 3 m/s. What is its kinetic energy?

  1. 6 J
  2. 9 J Correct answer.
  3. 3 J
  4. 18 J

Explanation

Kinetic energy is ½mv² = ½(2)(3)² = 9 J. The velocity is squared, so doubling speed quadruples the kinetic energy.

Question 6Classical Mechanics

A particle moves in a circle of radius r at constant speed v. What is the magnitude of its acceleration?

  1. Zero, because the speed is constant
  2. v/r
  3. vr
  4. v²/r Correct answer.

Explanation

Uniform circular motion has centripetal acceleration directed toward the center with magnitude v²/r. The speed is constant but the velocity direction changes, so acceleration is nonzero.

Question 7Classical Mechanics

A 5 kg block rests on a horizontal floor with coefficient of static friction 0.4 (take g = 10 m/s²). What is the minimum horizontal force needed to start it moving?

  1. 2 N
  2. 5 N
  3. 50 N
  4. 20 N Correct answer.

Explanation

The maximum static friction is μsN = μs mg = 0.4 × 5 × 10 = 20 N, so a horizontal force just above 20 N starts motion. The normal force equals the weight on a horizontal floor.

Question 8Classical Mechanics

A satellite is in a circular orbit of radius r around a planet of mass M. What is its orbital speed?

  1. GM/r²
  2. √(GM/r) Correct answer.
  3. √(2GM/r)
  4. √(GM/r³)

Explanation

Setting the gravitational force equal to the centripetal requirement gives GMm/r² = mv²/r, so v = √(GM/r). The speed decreases as the orbital radius grows.

Question 9Classical Mechanics

A 3 kg mass moving at 4 m/s collides head-on with a stationary 1 kg mass and they stick together. What is the speed of the combined mass after the collision?

  1. 1 m/s
  2. 4 m/s
  3. 3 m/s Correct answer.
  4. 2 m/s

Explanation

Momentum is conserved in a perfectly inelastic collision: (3)(4) = (3+1)v, giving v = 12/4 = 3 m/s. Kinetic energy is not conserved, but total momentum is.

Question 10Classical Mechanics

A spring of constant k = 200 N/m is compressed 0.1 m from equilibrium. How much elastic potential energy is stored?

  1. 10 J
  2. 1 J Correct answer.
  3. 2 J
  4. 20 J

Explanation

Elastic potential energy is ½kx² = ½(200)(0.1)² = ½(200)(0.01) = 1 J. The displacement is squared, so the energy grows rapidly with compression.

Question 11Classical Mechanics

A solid disk of mass M and radius R rolls without slipping down an incline. What fraction of its total kinetic energy is rotational?

  1. Two thirds
  2. One third Correct answer.
  3. One half
  4. One quarter

Explanation

For a solid disk I = ½MR², and rolling without slipping gives ω = v/R, so K_rot = ½Iω² = ¼Mv² while K_trans = ½Mv². The rotational fraction is (¼)/(¾) = ⅓.

Question 12Classical Mechanics

A particle moves under a conservative force. Which statement about its total mechanical energy is correct?

  1. It increases whenever the particle speeds up
  2. It is conserved, so it stays constant in time Correct answer.
  3. It is conserved only if the particle moves in a straight line
  4. It decreases because the force does negative work

Explanation

For a conservative force, work done is exactly the decrease in potential energy, so kinetic plus potential energy stays constant. The path shape and speed changes do not affect conservation.

Question 13Classical Mechanics

A uniform rod of length L and mass M is pivoted at one end and released from horizontal. What is its angular acceleration at the instant of release?

  1. 3g/(2L) Correct answer.
  2. 2g/L
  3. g/(2L)
  4. g/L

Explanation

The torque about the pivot is Mg(L/2), and the moment of inertia about the end is ML²/3. Thus α = τ/I = (MgL/2)/(ML²/3) = 3g/(2L).

Question 14Classical Mechanics

A simple pendulum has period T on Earth. On a planet where the gravitational acceleration is one quarter of Earth's, what is its period there?

  1. 4T
  2. 2T Correct answer.
  3. T/4
  4. T/2

Explanation

The period of a simple pendulum is T = 2π√(L/g), so it scales as 1/√g. Reducing g by a factor of 4 multiplies the period by √4 = 2.

Question 15Classical Mechanics

A 2 kg object moving at 6 m/s strikes a wall and rebounds at 4 m/s along the same line. What is the magnitude of the impulse delivered to the object?

  1. 8 N·s
  2. 4 N·s
  3. 20 N·s Correct answer.
  4. 24 N·s

Explanation

Taking the initial direction as positive, the change in momentum is m(v_f − v_i) = 2(−4 − 6) = −20 kg·m/s, so the impulse magnitude is 20 N·s. The rebound reverses the sign of the velocity.

Question 16Classical Mechanics

A block of mass m is attached to a spring of constant k on a frictionless horizontal surface and executes simple harmonic motion. What is the angular frequency of oscillation?

  1. √(k/m) Correct answer.
  2. k/m
  3. 2π√(m/k)
  4. √(m/k)

Explanation

For a mass-spring oscillator the equation of motion gives ω = √(k/m) in radians per second. The period is T = 2π/ω = 2π√(m/k).

Question 17Classical Mechanics

A particle moves in a central force field. Which quantity is conserved as a consequence of the force being central?

  1. The radial component of velocity
  2. Total mechanical energy only if the orbit is circular
  3. Angular momentum about the force center Correct answer.
  4. Linear momentum

Explanation

A central force exerts zero torque about the force center, so angular momentum about that point is conserved. This underlies Kepler's law of equal areas.

Question 18Classical Mechanics

A uniform solid sphere and a hoop of the same mass and radius are released from rest at the same height and roll without slipping down the same incline. Which reaches the bottom first?

  1. The hoop
  2. It depends on the incline angle
  3. The solid sphere Correct answer.
  4. They arrive at the same time

Explanation

Acceleration for rolling is a = g sin θ/(1 + I/(MR²)). The sphere has I/(MR²) = 2/5 and the hoop has 1, so the sphere has the larger acceleration and arrives first.

Question 19Classical Mechanics

A block of mass m slides down a rough incline of angle θ with coefficient of kinetic friction μk. What is its acceleration along the incline?

  1. g sin θ
  2. g(cos θ − μk sin θ)
  3. g(sin θ − μk cos θ) Correct answer.
  4. g(sin θ + μk cos θ)

Explanation

The driving component is mg sin θ and kinetic friction opposes motion with magnitude μk mg cos θ. Dividing the net force by mass gives a = g(sin θ − μk cos θ).

Question 20Classical Mechanics

A 1 kg mass moving east at 3 m/s collides with a 2 kg mass moving north at 3 m/s and they stick together. What is the speed of the combined mass?

  1. 3 m/s
  2. √5 m/s Correct answer.
  3. 1 m/s
  4. 2 m/s

Explanation

Momentum is conserved as a vector: p = (3, 6) kg·m/s, so |p| = √(9+36) = √45. Dividing by total mass 3 kg gives v = √45/3 = √5 m/s.

Question 21Classical Mechanics

A particle moves in one dimension under a potential U(x) = ½kx² with k > 0. Near the equilibrium at x = 0, what is the frequency of small oscillations for mass m?

  1. √(m/k)
  2. k/m
  3. √(k/m) Correct answer.
  4. 2π√(k/m)

Explanation

Near a minimum, the potential is harmonic with effective spring constant U''(0) = k, so ω = √(k/m). The equilibrium is stable because the second derivative is positive.

Question 22Classical Mechanics

A uniform rod of length L and mass M rotates about an axis through its center perpendicular to its length. What is its moment of inertia?

  1. ML²/2
  2. ML²/3
  3. ML²
  4. ML²/12 Correct answer.

Explanation

For a uniform rod about its center, I = ML²/12. By the parallel-axis theorem, the moment about one end is ML²/12 + M(L/2)² = ML²/3.

Question 23Classical Mechanics

A particle moves in a circular orbit under an inverse-square central force. How does its orbital period depend on the orbital radius r?

  1. T ∝ r^(3/2) Correct answer.
  2. T ∝ r²
  3. T ∝ r^(1/2)
  4. T ∝ r

Explanation

Circular orbital speed satisfies v² = GM/r, and T = 2πr/v = 2πr/√(GM/r) = 2π r^(3/2)/√(GM). This is Kepler's third law for circular orbits.

Question 24Classical Mechanics

A particle of mass m moves in a one-dimensional potential U(x) = A/x² − B/x with A, B > 0. At what radius is the circular orbit stable, and what is the angular frequency of small radial oscillations about it?

  1. r₀ = 2A/B, ω = B²/(2A²m)^(1/2) Correct answer.
  2. r₀ = A/B, ω = B²/(A²m)^(1/2)
  3. r₀ = 2A/B, ω = B/(2Am)^(1/2)
  4. r₀ = B/A, ω = B²/(2A²m)^(1/2)

Explanation

Setting the effective-force condition gives r₀ = 2A/B. Expanding the effective potential about this minimum yields ω = √(U''_eff(r₀)/m) = √(B⁴/(4A⁴m)) = B²/(2A²√m).

Question 25Classical Mechanics

A uniform chain of length L and mass M lies on a frictionless table with a length x hanging over the edge. What is the acceleration of the chain, and what is the condition for equilibrium?

  1. a = g(L−x)/L, equilibrium at x = L
  2. a = gx/L, equilibrium only at x = 0 Correct answer.
  3. a = gx/(L−x), equilibrium at x = L/2
  4. a = gx²/L², equilibrium at x = 0

Explanation

The hanging part has weight Mgx/L pulling the whole chain of mass M, so a = gx/L. Because the net force is nonzero for any x > 0, the only equilibrium is x = 0.

Question 26Classical Mechanics

A projectile is launched over level ground at speed v0 and angle θ above the horizontal. Neglecting air resistance, which expression gives its horizontal range?

  1. 2v0² sinθ/g
  2. v0² cos²θ/g
  3. v0² sin(2θ)/g Correct answer.
  4. v0² sin²θ/(2g)

Explanation

The time of flight is 2v0 sinθ/g and the horizontal velocity is constant at v0 cosθ, so the range is their product, 2v0² sinθcosθ/g, which equals v0² sin(2θ)/g. This is maximized at θ = 45° and vanishes at θ = 0° and 90°, as expected physically.

Question 27Classical Mechanics

A force F = (3i + 4j) N acts on a particle that undergoes a displacement d = (2i − 1j) m. How much work does the force do on the particle?

  1. 10 J
  2. 2 J Correct answer.
  3. 5 J
  4. 6 J

Explanation

Work is the dot product of force and displacement: (3)(2) + (4)(−1) = 6 − 4 = 2 J. Only the component of force along the displacement contributes, which is why the perpendicular j-component reduces rather than adds to the result.

Question 28Classical Mechanics

Two forces act at different points on a rigid body. Which condition is necessary for the body to be in static equilibrium?

  1. The net torque about every point must vanish and the net force must vanish. Correct answer.
  2. The net force must vanish, but the net torque may be nonzero about some points.
  3. The net force must vanish about the center of mass only.
  4. The net torque must vanish about the center of mass only.

Explanation

Static equilibrium requires both zero net force and zero net torque. If the net torque vanished about only one point, a couple could still rotate the body, so the torque condition must hold about any chosen point.

Question 29Classical Mechanics

A particle of mass m moves in one dimension in a potential U(x). Its total energy E is conserved. Which statement about its turning points is correct?

  1. They occur where the force −dU/dx is largest.
  2. They occur where the velocity is maximum.
  3. They occur where U(x) equals E, so the kinetic energy vanishes there. Correct answer.
  4. They occur where U(x) is a minimum.

Explanation

Since E = ½mv² + U(x), the particle can reach only positions where U(x) ≤ E. At the turning points U(x) = E, so the kinetic energy and hence speed are zero and the particle reverses direction.

Question 30Classical Mechanics

A uniform thin rod of mass M and length L is pivoted at one end about an axis perpendicular to the rod. A small blob of putty of mass m strikes the free end perpendicular to the rod and sticks to it. Which principle determines the angular speed just after the collision?

  1. Conservation of mechanical energy
  2. Conservation of angular momentum about the pivot Correct answer.
  3. Conservation of angular momentum about the rod's center of mass
  4. Conservation of linear momentum alone

Explanation

During the collision the pivot exerts an external impulse, so linear momentum is not conserved and energy is lost in the inelastic sticking. The pivot exerts no torque about itself, so angular momentum about the pivot is conserved and gives the post-collision angular speed.

Question 31Classical Mechanics

A particle moves in a plane with position vector r and velocity v. For a central force directed toward the origin, which quantity is conserved and why?

  1. The radial component of velocity, because the force is along r.
  2. Angular momentum r × mv, because the torque r × F vanishes. Correct answer.
  3. Linear momentum, because the force is radial.
  4. Kinetic energy, because central forces are always constant in magnitude.

Explanation

A central force is parallel to r, so the torque r × F is zero and angular momentum r × mv is constant. A consequence is that the motion stays in a plane perpendicular to the conserved angular momentum vector.

Question 32Classical Mechanics

Two identical springs, each of spring constant k, are connected in series and support a mass m. What is the angular frequency of small vertical oscillations?

  1. √(k/m)
  2. 2√(k/m)
  3. √(2k/m)
  4. √(k/(2m)) Correct answer.

Explanation

Springs in series share the load, so the equivalent constant is k_eq = k/2. The angular frequency is √(k_eq/m) = √(k/(2m)), which is lower than for a single spring because the combination is softer.

Question 33Classical Mechanics

A uniform solid cylinder of mass M and radius R rolls without slipping on a horizontal surface. What is the ratio of its rotational kinetic energy to its translational kinetic energy?

  1. 1/2 Correct answer.
  2. 1
  3. 2
  4. 1/4

Explanation

For a solid cylinder I = ½MR², and rolling without slipping gives ω = v/R. The rotational energy ½Iω² = ¼Mv² and the translational energy ½Mv², so the ratio is ¼/(½) = 1/2.

Question 34Classical Mechanics

A particle moves in a circular orbit of radius r under an inverse-square attractive force of magnitude k/r². What is the total mechanical energy of the particle?

  1. −k/(2r) Correct answer.
  2. k/(2r)
  3. −k/r
  4. k/r

Explanation

For a circular orbit under F = k/r², the centripetal condition gives mv² = k/r, so the kinetic energy is k/(2r). The potential energy is −k/r, so the total energy is k/(2r) − k/r = −k/(2r).

Question 35Classical Mechanics

A block slides down a frictionless track and enters a vertical circular loop of radius R, starting from rest at a height h above the bottom of the loop. What minimum height h allows the block to just maintain contact at the top of the loop?

  1. 2R
  2. 5R/2 Correct answer.
  3. 3R
  4. R

Explanation

At the top, contact is just maintained when gravity alone provides the centripetal force: mg = mv²/R, giving v² = gR. Energy conservation from height h to the top (height 2R) gives mgh = mg(2R) + ½m(gR), so h = 5R/2.

Question 36Classical Mechanics

A uniform rod of length L and mass M is pivoted at one end and hangs vertically at rest. A horizontal impulse J is applied at the free end. What is the angular speed immediately after the impulse?

  1. 3J/(ML) Correct answer.
  2. 6J/(ML)
  3. J/(ML²)
  4. J/(ML)

Explanation

The angular impulse about the pivot is JL, and the moment of inertia about the end is I = ML²/3. Setting JL = Iω = (ML²/3)ω gives ω = 3J/(ML).

Question 37Classical Mechanics

A particle moves in a central force field with potential U(r) = −k/r. Which statement about its orbit is correct for negative total energy?

  1. The orbit is a straight line through the force center.
  2. The orbit is a conic section with the force center at one focus. Correct answer.
  3. The orbit is always a circle centered on the force center.
  4. The orbit is a spiral that eventually escapes.

Explanation

The inverse-square force produces Kepler-type motion, in which bound orbits (E < 0) are ellipses with the force center at one focus. Circular and radial orbits are special cases corresponding to particular values of angular momentum and energy.

Question 38Classical Mechanics

A rocket moves in deep space with no external forces. It ejects fuel at constant exhaust speed u relative to the rocket. If the rocket's mass decreases from m0 to m, what is the change in its speed?

  1. u ln(m0/m) Correct answer.
  2. u(m0 − m)/m
  3. u(m0/m − 1)²
  4. u(m0 − m)/m0

Explanation

The rocket equation follows from momentum conservation with variable mass: dv = u dm/m integrated gives Δv = u ln(m0/m). The logarithmic form reflects that later fuel is ejected from a lighter, faster rocket and is more effective.

Question 39Classical Mechanics

A particle moves in a one-dimensional potential U(x) = −A/x + B/x² with A, B > 0. What is the angular frequency of small oscillations about the stable equilibrium?

  1. √(A⁴/(8B³m)) Correct answer.
  2. √(A/(Bm))
  3. √(A²/(2Bm))
  4. √(2A²/(B³m))

Explanation

Setting dU/dx = A/x² − 2B/x³ = 0 gives x0 = 2B/A. The second derivative is U'' = −2A/x³ + 6B/x⁴, which at x0 equals A⁴/(8B³). The oscillation frequency is √(U''/m) = √(A⁴/(8B³m)).

Question 40Classical Mechanics

A uniform chain of length L and mass M lies on a frictionless table with a length x hanging over the edge. The chain is released from rest. What is the acceleration of the chain when the hanging length is x?

  1. g(L − x)/L
  2. gx²/L²
  3. g/2
  4. gx/L Correct answer.

Explanation

The driving force is the weight of the hanging part, (M/L)xg, and the total mass M is accelerated. Newton's second law gives Ma = (M/L)xg, so a = gx/L, which grows as more chain hangs over the edge.

Question 41Electromagnetism

A point charge +Q is placed at the center of an uncharged, isolated conducting spherical shell of inner radius a and outer radius b. What is the total electric field at a point a distance r from the center, where a < r < b?

  1. Q/(4πε₀r²) directed radially outward
  2. Zero Correct answer.
  3. Q/(4πε₀r²) directed radially inward
  4. Q/(4πε₀a²) directed radially outward

Explanation

Inside the conducting material the electrostatic field must vanish in equilibrium, otherwise free charges would keep moving. The inner surface acquires charge −Q and the outer surface +Q, but within the metal itself the net field is zero.

Question 42Electromagnetism

Two point charges, +q and −q, are separated by a distance d. At the midpoint of the line joining them, what are the electric field and the electric potential (taking V = 0 at infinity)?

  1. E = 2kq/d² pointing toward +q, V = 0
  2. E = 8kq/d² pointing toward −q, V = 0 Correct answer.
  3. E = 0, V = 0
  4. E = 0, V = 2kq/(d/2)

Explanation

At the midpoint each charge contributes a field of magnitude kq/(d/2)² = 4kq/d², and both point toward the negative charge, so they add to 8kq/d². The potentials +kq/(d/2) and −kq/(d/2) cancel, giving V = 0.

Question 43Electromagnetism

A parallel-plate capacitor has plate area A and separation d. A dielectric slab of constant κ completely fills the gap. Compared with the air-filled capacitor, the capacitance is

  1. unchanged
  2. multiplied by κ²
  3. divided by κ
  4. multiplied by κ Correct answer.

Explanation

For fixed charge the dielectric reduces the field by the factor κ, so the potential difference drops by κ and C = Q/V rises by κ. Thus C = κε₀A/d for a fully filled capacitor.

Question 44Electromagnetism

A steady current I flows through a straight wire of length L that lies along the x-axis. What is the magnitude of the magnetic field produced at a perpendicular distance r from the wire's midpoint, in the limit L ≫ r?

  1. μ₀I/(4πr)
  2. μ₀I/(2r)
  3. μ₀IL/(2πr²)
  4. μ₀I/(2πr) Correct answer.

Explanation

Ampère's law applied to a circular loop of radius r around a long straight wire gives B·2πr = μ₀I, hence B = μ₀I/(2πr). The field lines are circles centered on the wire.

Question 45Electromagnetism

A charged particle moves with velocity v in a region containing a uniform magnetic field B. Which statement about the magnetic force on the particle is correct?

  1. It is parallel to B and changes the particle's kinetic energy.
  2. It is perpendicular to v but parallel to B, doing work only when v is perpendicular to B.
  3. It is perpendicular to both v and B and does no work on the particle. Correct answer.
  4. It is parallel to v and does work on the particle.

Explanation

The Lorentz magnetic force F = qv × B is by construction perpendicular to both v and B. Because it is always perpendicular to the velocity, it changes direction but never the speed, so it does zero work.

Question 46Electromagnetism

Two identical resistors of resistance R are connected in parallel across an ideal battery of emf V. What is the total power dissipated by the pair?

  1. 2V²/R Correct answer.
  2. V²/(2R)
  3. V²/R
  4. V²/(4R)

Explanation

Two resistors R in parallel give an equivalent resistance R/2. The power delivered by the battery is V²/(R/2) = 2V²/R, which also equals V²/R dissipated in each branch.

Question 47Electromagnetism

A uniform magnetic field B points out of the page. A square loop of wire lies in the plane of the page. The loop is being rotated about an axis in its plane at angular speed ω. At the instant the loop's normal is parallel to B, the induced emf is

  1. equal to BAω/2
  2. zero Correct answer.
  3. maximum
  4. equal to BAω

Explanation

The flux is Φ = BA cos θ, so the emf is −dΦ/dt = BAω sin θ. When the normal is parallel to B, θ = 0 and sin θ = 0, so the induced emf vanishes at that instant.

Question 48Electromagnetism

Which of the following is the correct statement of Gauss's law for the electric field in integral form?

  1. ∮E·dA = Q_enclosed/ε₀ over any closed surface Correct answer.
  2. ∮E·dl = −dΦ_B/dt over any closed loop
  3. ∮E·dA = Q_enclosed·ε₀ over any closed surface
  4. ∮B·dA = μ₀I_enclosed over any closed surface

Explanation

Gauss's law states that the electric flux through any closed surface equals the enclosed charge divided by ε₀. The surface may be chosen freely, which is what makes the law useful for symmetric charge distributions.

Question 49Electromagnetism

A long solenoid of n turns per unit length carries current I. What is the magnetic field inside, far from the ends?

  1. μ₀nI, but radially outward
  2. μ₀I/(2πn), uniform and axial
  3. μ₀nI, uniform and axial Correct answer.
  4. μ₀nI/2, uniform and axial

Explanation

Applying Ampère's law to a rectangular loop with one side inside the solenoid gives B·ℓ = μ₀nℓI, so B = μ₀nI. The field is essentially uniform and directed along the axis inside a long solenoid.

Question 50Electromagnetism

A charge q moves in a circle of radius r with constant speed v under the influence of a uniform magnetic field perpendicular to the plane of the circle. The cyclotron frequency of the motion is

  1. qB/(2πm) Correct answer.
  2. qvB/(2πr)
  3. qB/(2πr)
  4. qvB/(2πm)

Explanation

Setting qvB = mv²/r gives angular frequency ω = qB/m, so the frequency is f = qB/(2πm). Notably, the cyclotron frequency is independent of speed and radius for nonrelativistic motion.