Examination Style Questions 7 from Oscillations note

The apparatus of Fig. 13.16 is used to demonstrate forced vibrations and resonance. A 50 g mass is suspended from the spring, which has a spring constant of 7.9 N m⁻¹.

a)  Calculate the resonant frequency f₀ of the system.

b)  A student suggests that resonance should also be observed at a frequency of 2 f₀. Discuss this suggestion.

Solution:

a)

b) If the driving frequency is increased further, the amplitude of oscillation of the mass decreases.

Reference: Examination Style Questions 7 from note

9702/Oct Nov/12/2017/Q7

The velocity-time graph for a train starting at one station and stopping at the next is shown.

Another train has double the acceleration but the same maximum speed and the same

deceleration.

Which velocity-time graph, on the same scale, shows the motion of this train between the same

stations?

Solution:

Answer: B

From option A to D, we know that all of the acceleration become double from beginning, after that same maximum speed and at the end same deceleration.

However, the question had mentions the same station for the train (which mean same displacement).

The displacement of the train = velocity x time
= the area of trapezium of the graph.
= 1/2 (11+5)(8)
= 64m

For option A, the displacement = 1/2 (9+5)(8) = 56m

For option B, the displacement = 1/2 (10+6)(8) = 64m

For option C, the displacement = 1/2 (8+4)(8) = 48m

For option D, the displacement = 1/2 (11+7)(8) = 72m

Reference: PYQ - Oct/Nov 2017 Paper 12 Q7

9702/Oct Nov/11/2015/Q7

One of the equations of uniformly accelerated motion is shown.

s = ut + ½ at²

Apparatus is arranged to record the time t taken for a marble to fall between two light gates

connected to timers. The marble touches the stop before it is released. The vertical distance s

between the light gates is measured.

Which graph does not show a correct relationship when light gate 2 moves up to light gate 1
which is fixed?

Solution:
Answer: C

Choice A: from the formula s = ut + ½ at² , we know s ∝ t²  , so choice A is correct.

Choice B: we can derive the formula of s/t = u + ½ at. s/t ∝ t , so choice B is correct.

Choice D: The gravitational acceleration is always constant. So choice D is correct.

Choice C:
We can derive the formula of
ut = s – ½ at²

u = s/t – ½ at

So u = 0 at all the time is doesn't make sense.

Reference: PYQ - Oct/Nov 2015 Paper 11 Q7

9702/May Jun/12/2015/Q7

In an experiment to determine the acceleration of free fall g, a ball bearing is held by an
electromagnet. When the current to the electromagnet is switched off, a clock starts and the ball
bearing falls. After falling a distance h, the ball bearing strikes a switch to stop the clock which
measures the time t of the fall.

If systematic errors cause t and h to be measured incorrectly, which error must cause g to
appear greater than 9.81 m s–2?

A. h measured as being smaller than it actually is and t is measured correctly
B. h measured as being smaller than it actually is and t measured as being larger than it actually is
C. h measured as being larger than it actually is and t measured as being larger than it actually is
D. h is measured correctly and t measured as being smaller than it actually is

Solution:
Answer: D.

Gravitational acceleration (symbolized g) is an expression used in physics to indicate the intensity of a gravitational field. It is expressed in meters per second squared (m/s 2 ).
g = m/s ²

Let h = m, t = s .

Consider choice A:
m decrease , t maintain
the gravitational acceleration appear lesser than 9.81 m/s ²

Consider choice B:
m decrease , t increase
the gravitational acceleration appear lesser than 9.81 m/s ²

Consider choice C:
m increase , t increase
the gravitational acceleration appear lesser than 9.81 m/s ² due to s is power 2.

Consider choice D:
m maintain, t decrease
the gravitational acceleration appear greater than 9.81 m/s ².

Reference: PYQ - May/Jun 2015 Paper 12 Q7