Unit 7 · Topic 7.3 Beta

Representing and Analyzing SHM

SHM is described by x = A cos(2πft) or A sin(2πft); on the graphs, speed is greatest and acceleration zero at equilibrium, speed zero and acceleration greatest at the turning points, and the period is independent of the amplitude.

Practice 1: Creating RepresentationsPractice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

A seismograph, a heart monitor and a motion detector all draw a wave-like trace across a screen. When the trace comes from a mass on a spring, you can read everything about the motion from it: how far the mass swings, how long each swing takes, and where it moves fastest.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. On a position-time graph, the slope gives

  1. the velocity
  2. the acceleration
  3. the displacement
  4. the distance traveled
Show the answer

Velocity is the rate of change of position, the slope of x against t.

  • Correct: the velocity:
  • the acceleration:
  • the displacement:
  • the distance traveled:

2. What is cos 0?

  1. 1
  2. 0
  3. −1
  4. 0.5
Show the answer

cos 0 = 1 and sin 0 = 0.

  • Correct: 1:
  • 0:
  • −1:
  • 0.5:

3. A block on a spring is pulled twice as far. Its period

  1. stays the same
  2. doubles
  3. halves
  4. grows by √2
Show the answer

T = 2π√(m/k) has no amplitude in it.

  • Correct: stays the same:
  • doubles:
  • halves:
  • grows by √2:

Part 4 · See it

See it first

Position, velocity and acceleration against time over one period for SHM starting at x = +A. Position is a cosine; velocity is zero at the turning points and largest at x = 0; acceleration is always opposite to the position, largest at the turning points and zero at equilibrium.
Position, velocity and acceleration over one period, lined up in time. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A restoring force F = −kx acts on the objectits acceleration is always opposite to its displacement, so the a-t graph is the x-t graph flipped
  2. The object turns around at x = ±Aits velocity is zero there: the x-t graph is flat at its peaks and troughs
  3. The object passes x = 0 with no force to slow it yetits speed is greatest there: the x-t graph is steepest where it crosses zero
  4. Doubling the amplitude doubles the force and the speed at each matching pointthe cycle takes the same time: the period does not depend on the amplitude

Part 6 · Key ideas

Key ideas

  • Position in SHM is x = A cos(2πft) if the object starts at +A, or x = A sin(2πft) if it starts at equilibrium moving in +x. Use radians.
  • On an x-t graph: the height of a peak is A; the time from peak to peak is T; the slope is the velocity.
  • Speed is greatest at x = 0 and zero at x = ±A. Acceleration is zero at x = 0 and largest at x = ±A, always pointing toward equilibrium.
  • The period does not depend on the amplitude: a bigger swing has more distance to cover and covers it faster.

Part 7 · Misconception

A common mistake

The wrong idea: The object moves fastest at the turning points, because that is where the force on it is largest.

What actually happens: A large force means a large acceleration, not a large speed. At the turning points the object is momentarily at rest; it is fastest at equilibrium, where the force is zero.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Position of a cart on a spring

A motion detector records the position of a cart oscillating on a horizontal spring on a frictionless track. Position is measured from equilibrium; + is to the right.

-0.2-0.15-0.1-0.0500.050.10.150.200.20.40.60.811.21.41.6Time (s)Position (m)
Data table
Time (s)Cart
00.15
0.050.139
0.10.106
0.150.057
0.20
0.25-0.057
0.3-0.106
0.35-0.139
0.4-0.15
0.45-0.139
0.5-0.106
0.55-0.057
0.60
0.650.057
0.70.106
0.750.139
0.80.15
0.850.139
0.90.106
0.950.057
10
1.05-0.057
1.1-0.106
1.15-0.139
1.2-0.15
1.25-0.139
1.3-0.106
1.35-0.057
1.40
1.450.057
1.50.106
1.550.139
1.60.15

1. What is the period of the cart's motion? Give the value and its unit.

Type a number and its unit.

Show the answer

The cart is at +0.15 m at t = 0 and next at t = 0.80 s, so T = 0.80 s. Half a period (0.40 s) only takes it to the other side.

  • Answer: 0.80 s

2. At which time is the cart moving fastest?

  1. t = 0.20 s
  2. t = 0.40 s
  3. t = 0.80 s
  4. t = 0.30 s
Show the answer

Speed is the steepness of the position-time graph. The graph is steepest where it crosses x = 0, at t = 0.20 s (and 0.60 s, 1.00 s...). The cart passes equilibrium at top speed.

  • Correct: t = 0.20 s: Right: the slope is steepest where x = 0.
  • t = 0.40 s: At 0.40 s the cart is at −0.15 m, a turning point: the slope is zero and the cart is momentarily at rest.
  • t = 0.80 s: At 0.80 s the cart is back at +0.15 m, a turning point with zero speed.
  • t = 0.30 s: At 0.30 s the cart is between equilibrium and a turning point; it is still moving, but slower than at 0.20 s.

3. What are the directions of the cart's velocity and acceleration at t = 0.50 s?

  1. Velocity +, acceleration +
  2. Velocity −, acceleration +
  3. Velocity +, acceleration −
  4. Velocity −, acceleration −
Show the answer

At 0.50 s the cart is below zero (x < 0) and the graph is rising, so v is +. The acceleration always points toward equilibrium, opposite to x; x is negative, so a is +.

  • Correct: Velocity +, acceleration +: Right: x < 0 and rising, so v > 0 and a = −(k/m)x > 0.
  • Velocity −, acceleration +: The graph is rising at 0.50 s (it left −0.15 m at 0.40 s), so the velocity is positive.
  • Velocity +, acceleration −: Acceleration is opposite to the displacement; with x negative, a is positive.
  • Velocity −, acceleration −: Both signs are off: the graph is rising at 0.50 s (v > 0), and with x below zero the acceleration points toward equilibrium (a > 0).

4. The cart is stopped and released again from x = +0.075 m. Which graph feature changes, compared with the graph shown?

  1. The peaks are half as high, and the time between them is unchanged
  2. The peaks are half as high, and the time between them is halved
  3. The peaks are the same height, and the time between them is halved
  4. The peaks are half as high, and the time between them is doubled
Show the answer

Releasing from 0.075 m makes the amplitude 0.075 m, half of 0.15 m. The period T = 2π√(m/k) does not depend on the amplitude, so the peaks stay 0.80 s apart.

  • Correct: The peaks are half as high, and the time between them is unchanged: Right: smaller amplitude, same period.
  • The peaks are half as high, and the time between them is halved: A smaller amplitude does not shorten the period: less distance, but proportionally less force and speed.
  • The peaks are the same height, and the time between them is halved: The release point sets the amplitude, so the peaks must be lower.
  • The peaks are half as high, and the time between them is doubled: The period depends on m and k only; amplitude does not lengthen it.

5. Two identical blocks on identical springs are released from rest at the same moment, block 1 from 3.0 cm and block 2 from 6.0 cm. Which block returns to its release point first?

  1. They return at the same time
  2. Block 1, because it has a shorter distance to travel
  3. Block 2, because it is pulled back by a larger force
  4. Block 1, because its smaller force means less acceleration
Show the answer

The period T = 2π√(m/k) is the same for both: block 2 travels twice as far but feels twice the force at each matching point, so it moves twice as fast. The two effects cancel.

  • Correct: They return at the same time: Right: the period does not depend on the amplitude.
  • Block 1, because it has a shorter distance to travel: Block 1 does travel less distance, but it also moves more slowly; the times match.
  • Block 2, because it is pulled back by a larger force: Block 2 does feel a larger force, but it also has farther to go; the times match.
  • Block 1, because its smaller force means less acceleration: Less acceleration slows block 1 down, which works against it, not for it; the times still match.

6. A block on a spring is released from rest at x = +A at t = 0. Which equation describes its position?

  1. x = A cos(2πft)
  2. x = A sin(2πft)
  3. x = −A cos(2πft)
  4. x = A sin(2πft) + A
Show the answer

At t = 0 the block must be at +A: cos 0 = 1 gives x = A. Both sine and cosine describe SHM; the starting point decides which one fits.

  • Correct: x = A cos(2πft): Right: cosine starts at its largest value.
  • x = A sin(2πft): This starts at x = 0, which fits a block that starts at equilibrium.
  • x = −A cos(2πft): This starts at x = −A, the other turning point.
  • x = A sin(2πft) + A: This oscillates between 0 and 2A, so its center is not the equilibrium position.

Part 9 · Summary

Summary

SHM is described by x = A cos(2πft) or A sin(2πft); on the graphs, speed is greatest and acceleration zero at equilibrium, speed zero and acceleration greatest at the turning points, and the period is independent of the amplitude.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections