PHY26041/102/0/2026 iQ Level: 0793226427
UNISA
PHY2606 Assignment 6 full solutions 2026
Assignment 6
Due Date: 11th September 2026 at 09:00
Assignment Unique number: 209220
Instruction
- This assignment consists of eleven (11) questions.. each question
with Notes on how and why correct option is correct. At the bottom of the document
there is a list of question and correct options too.
, Question 1
From a position-versus-time graph for SHM:
(a) Acceleration is zero
Acceleration 𝑎 = 0 when 𝑥 = 0 (since 𝑎 = −𝜔2 𝑥 for SHM).
→ At equilibrium points, where displacement 𝑥 = 0. Label these points on the graph.
(b) Kinetic energy is maximum
Kinetic energy maximum at equilibrium (𝑥 = 0) → same points as (a).
(c) Potential energy is maximum
Potential energy maximum when 𝑥 is maximum (amplitude).
→ At extreme positions.
So the reasoning is same graph-based answer:
(a) points where 𝑥 = 0
(b) same points as (a)
(c) points where ∣ 𝑥 ∣ max.
Final answer: But since no exact graph given, notation is:
(a) 𝑥 = 0 points, (b) 𝑥 = 0 points, (c) ∣ 𝑥 ∣= 𝐴 points.
Question 2
Again, from same graph (but Figure 2):
In SHM, 𝑥-𝑡 graph is 𝐴cos(𝜔𝑡 + 𝜙) or 𝐴sin.
From 𝑥(𝑡):
Velocity 𝑣(𝑡) = −𝐴𝜔sin(𝜔𝑡 + 𝜙)
Acceleration 𝑎(𝑡) = −𝐴𝜔2 cos(𝜔𝑡 + 𝜙)
Check sign combinations:
(a) 𝑣 < 0, 𝑎 < 0
𝑎 = −𝜔2 𝑥, so 𝑎 < 0 ⟺ 𝑥 > 0.
𝑣 < 0 = slope negative. So 𝑥 > 0 and slope negative → after max 𝑥 going toward 0.
(b) 𝑣 > 0, 𝑎 < 0
𝑎 < 0 ⟺ 𝑥 > 0; 𝑣 > 0 means slope positive → between 0 and max positive → for 𝑥 > 0 and
slope positive → from 0 to max.
(c) 𝑣 < 0, 𝑎 > 0
𝑎 > 0 ⟺ 𝑥 < 0; 𝑣 < 0 is slope negative → in 𝑥 < 0 region, moving more negative → after -A
toward 0? Wait, if x<0 and slope negative, means moving away from 0 towards negative max.
Yes, so in x<0 region, slope negative = from -A to 0? Actually if x<0 and velocity negative,
2
UNISA
PHY2606 Assignment 6 full solutions 2026
Assignment 6
Due Date: 11th September 2026 at 09:00
Assignment Unique number: 209220
Instruction
- This assignment consists of eleven (11) questions.. each question
with Notes on how and why correct option is correct. At the bottom of the document
there is a list of question and correct options too.
, Question 1
From a position-versus-time graph for SHM:
(a) Acceleration is zero
Acceleration 𝑎 = 0 when 𝑥 = 0 (since 𝑎 = −𝜔2 𝑥 for SHM).
→ At equilibrium points, where displacement 𝑥 = 0. Label these points on the graph.
(b) Kinetic energy is maximum
Kinetic energy maximum at equilibrium (𝑥 = 0) → same points as (a).
(c) Potential energy is maximum
Potential energy maximum when 𝑥 is maximum (amplitude).
→ At extreme positions.
So the reasoning is same graph-based answer:
(a) points where 𝑥 = 0
(b) same points as (a)
(c) points where ∣ 𝑥 ∣ max.
Final answer: But since no exact graph given, notation is:
(a) 𝑥 = 0 points, (b) 𝑥 = 0 points, (c) ∣ 𝑥 ∣= 𝐴 points.
Question 2
Again, from same graph (but Figure 2):
In SHM, 𝑥-𝑡 graph is 𝐴cos(𝜔𝑡 + 𝜙) or 𝐴sin.
From 𝑥(𝑡):
Velocity 𝑣(𝑡) = −𝐴𝜔sin(𝜔𝑡 + 𝜙)
Acceleration 𝑎(𝑡) = −𝐴𝜔2 cos(𝜔𝑡 + 𝜙)
Check sign combinations:
(a) 𝑣 < 0, 𝑎 < 0
𝑎 = −𝜔2 𝑥, so 𝑎 < 0 ⟺ 𝑥 > 0.
𝑣 < 0 = slope negative. So 𝑥 > 0 and slope negative → after max 𝑥 going toward 0.
(b) 𝑣 > 0, 𝑎 < 0
𝑎 < 0 ⟺ 𝑥 > 0; 𝑣 > 0 means slope positive → between 0 and max positive → for 𝑥 > 0 and
slope positive → from 0 to max.
(c) 𝑣 < 0, 𝑎 > 0
𝑎 > 0 ⟺ 𝑥 < 0; 𝑣 < 0 is slope negative → in 𝑥 < 0 region, moving more negative → after -A
toward 0? Wait, if x<0 and slope negative, means moving away from 0 towards negative max.
Yes, so in x<0 region, slope negative = from -A to 0? Actually if x<0 and velocity negative,
2
