Physics

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Answer (Section A) Explanation
Question Answer
1. A = − 2

2. A = −(2 )2
�
= 1
2

= 1 � 0.6
2 0.02

f = 0.87 Hz

2 + = 0
⇒ 2

2 −
⇒ 2 =

⇒ = − 2

∴ 2 = = 2


⇒ = 2
√

Frequency ⇒ = 2
3. A
⇒ =
2

⇒ = 100 = 15.9
2

Kinetic energy

= 1 � � 2 − 2�2
2

= 1 (2) �100�52 − 32�2 = 160
2

4. D

1

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5. B

6. D f ' =  v + u0  f =  330 + 30 1000 = 1161Hz
v − u0  330 − 20 

7. B

8. C
D

9.

4. D.
10. D

11. D

2

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C
12.

13. D
D

14.

15. D

ANSWER PAPER SECTION B

Question Answer Mark
2m
Structure
16 1m
3
a) One difference and one similarity
b) i)
light wave sound wave

difference No need medium to Need medium to

propagate propagate

Can be polarized Cannot be polarized

similarity Both are progressive waves

Wave length ,λ = 1

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b) ii) Frequency, = = 10 = 10 Hz 4m
17(a) λ 1 2m
= 2 = 20

= sin( − 2
)
–ve sign → move in positive –x direction
2
= 0.5 sin( 20 − 1)

= 0.5 sin( 20 − 2 )

(b) 2m
(c) 3m

Essay 18

18(a) The motion of an object in which the acceleration is directly propotional to the
displacement from the displacement from the equilibrium position and is always
directed to the equilibrium position.

4

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= 0.70 cos(5 + ∅)

18(b)(i) Velocity , = (0.7)(5) sin(5 + ∅)
18(b)(ii)
18(b)(iii) = 3.5 sin (5t + ∅)
Maximum velocity = 3.5 ms-1
18(b)(iv)
Acceleration, = −(3.5)(5)cos (5 + ∅)
18(c)
= - 17.7 cos (5t + ∅)
Maximum acceleration = 17.7 ms-2

When t = 0, v = -0.40 ms-1

= −3.5 sin(5 + ∅)

-0.40 = -3.5 sin ∅
∅ = 6.56°

If the bob is displaced and then released, the displacement of the bob after a time t is

given by = 0 cos

Kinetic energy = potential energy 1
1 2
2 2( 2 − 2) = 2 2

= √2 = 0.707 0

When x = 0 cos = 0
√2 4

= =
8
4 = 4(2 ) =

19 Energy transfer per second (Power) per unit cross sectional area 1
(a-i) 5

Or I = E/A

E = energy (transfer) per second

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A = cross sectional area/ named area

a-ii The alternate rise and fall of sound level/ intensity due to the superposition of two
waves with slightly difference frequency

b-i

b-ii

c-i
c-ii

6

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c-iii
20 (a)

b-i
b-ii

b-iii

7

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c-i

c-ii

8