Displacement, Velocity and Acceleration in SHM

Displacement, Velocity and Acceleration in SHM

What is it?

For x = A cos(omega t): v = -A omega sin(omega t), a = -A omega^2 cos(omega t) = -omega^2 x. At mean (x = 0): |v| max = A omega, a = 0. At extremes (|x| = A): v = 0, |a| max = A omega^2. Energy form: v = +/- omega sqrt(A^2 - x^2). Signs show direction; exams often want magnitudes at named points.

In simple words

• For x = A cos(omega t): v = -A omega sin(omega t), a = -A omega^2 cos(omega t) = -omega^2 x.
• From X(T): Differentiate once for v, twice for a.
• Special Positions: Table: position | v | a.
• Exam reminder: Maximum speed is at the centre, not at maximum displacement.

Important terms

DISPLACEMENT, VELOCITY AND ACCELERATION IN SHM: For x = A cos(omega t): v = -A omega sin(omega t), a = -A omega^2 cos(omega t) = -omega^2 x.

FROM x(t): Differentiate once for v, twice for a.

SPECIAL POSITIONS: Table: position | v | a.

What you should learn

• Explain Displacement, Velocity And Acceleration In SHM in easy language and identify its main physical principle.
• Apply the key relation or evidence for Displacement, Velocity And Acceleration In SHM to a Class 12 conceptual or numerical question.
• Present an FBISE-style answer with correct reasoning, symbols, units, and a clear final statement.

Examples

• Cosine or sine: same physics, different t = 0
• Show the a = -omega^2 x line
• Units m, m/s, m/s^2

FROM x(t)

Differentiate once for v, twice for a. The second derivative must satisfy a = -omega^2 x to prove SHM.

SPECIAL POSITIONS

Table: position | v | a. Mean: fastest, a = 0. End: stop, largest |a| toward centre.

Full school-level explanation

For x = A cos(omega t): v = -A omega sin(omega t), a = -A omega^2 cos(omega t) = -omega^2 x. At mean (x = 0): |v| max = A omega, a = 0. At extremes (|x| = A): v = 0, |a| max = A omega^2. Energy form: v = +/- omega sqrt(A^2 - x^2). Signs show direction; exams often want magnitudes at named points.

Kinematics, then checkpoints:

Overview — Displacement, Velocity And Acceleration In SHM

For x = A cos(omega t): v = -A omega sin(omega t), a = -A omega^2 cos(omega t) = -omega^2 x. At mean (x = 0): |v| max = A omega, a = 0. At extremes (|x| = A): v = 0, |a| max = A omega^2. Energy form: v = +/- omega sqrt(A^2 - x^2). Signs show direction; exams often want magnitudes at named points.

From X(T)

Differentiate once for v, twice for a. The second derivative must satisfy a = -omega^2 x to prove SHM. Examples: Cosine or sine: same physics, different t = 0; Show the a = -omega^2 x line; Units m, m/s, m/s^2.

Special Positions

Table: position | v | a. Mean: fastest, a = 0. End: stop, largest |a| toward centre. Examples: Do not say max v at max x; Classic trap; |x| <= A always.

Worked or exam application

A = 0.04 m, T = 0.50 s. omega = 4 pi. v_max = 0.04 x 4 pi ~ 0.50 m s^-1.

FROM x(t)

Differentiate once for v, twice for a. The second derivative must satisfy a = -omega^2 x to prove SHM.

Remember

• dx/dt then d2x/dt2
• proves SHM

Examples

• Cosine or sine: same physics, different t = 0
• Show the a = -omega^2 x line
• Units m, m/s, m/s^2

SPECIAL POSITIONS

Table: position | v | a. Mean: fastest, a = 0. End: stop, largest |a| toward centre.

Remember

• table of extremes
• restoring a

Examples

• Do not say max v at max x
• Classic trap
• |x| <= A always

Remember

• v_max = A omega at mean
• a_max = A omega² at extremes
• v = omega sqrt(A² - x²)
• v_max = A omega
• mean = fastest
• dx/dt then d2x/dt2
• proves SHM
• table of extremes

How the ideas connect

A = 0.04 m, T = 0.50 s. omega = 4 pi. v_max = 0.04 x 4 pi ~ 0.50 m s^-1.

Deeper understanding

Go beyond memorising Displacement, Velocity and Acceleration in SHM. Connect the definition to reasons, relationships, exceptions, and exam wording.

How the ideas connect

A = 0.04 m, T = 0.50 s. omega = 4 pi. v_max = 0.04 x 4 pi ~ 0.50 m s^-1.

Common mistakes

Exam focus

Core Idea

From X(T)

Special Positions

Apply It

Exam focus

Quick revision

• v_max = A omega at mean
• a_max = A omega² at extremes
• v = omega sqrt(A² - x²)
• v_max = A omega
• DISPLACEMENT, VELOCITY AND ACCELERATION IN SHM: For x = A cos(omega t): v = -A omega sin(omega t), a = -A omega^2 cos(omega t) = -omega^2 x.
• FROM x(t): Differentiate once for v, twice for a.

Maximum speed is at the centre, not at maximum displacement.

For long answers on Displacement, Velocity and Acceleration in SHM: define clearly → explain the main parts → give one example → state one exam caution.