The diagram shows a stationary wave, at time t = 0, that has been set up on a string fixed between points P and S.

Question 22

The diagram shows a stationary wave, at time t = 0, that has been set up on a string fixed between points P and S.

The nodes of the stationary wave occur on the string at P, Q, R and S. Point X is moving down at time t = 0. The points on the string vibrate with time period T and maximum amplitude 2 cm.

The displacement s is positive in the upward direction.

Which graph best shows the variation with t of the displacement s of point Y on the string?

Reference: Past Exam Paper – June 2018 Paper 12 Q28

Solution:

Answer: A.

Loops would be formed between the nodes PQ, QR and RS.

Vibrations in adjacent loops are out of phase. That is, if the vibration is downwards at X (in the PQ loop), then vibrations in the loop QR would be upwards (out of phase with the loop PQ).

So, vibrations in loop RS would also be out of phase with loop QR as they are adjacent. Since vibrations in loop QR is upwards, it means that vibration is loop RS will be downwards.

So, the vibrations at points X and Y are in phase. The pattern of movement at Y will therefore be the same as at X. The displacement at Y will also be downwards – it has a negative displacement. [C and D are incorrect]

However, the maximum amplitude at Y will be lower than at X because X is at an antinode and Y is not. [A is correct]

The diagram shows two complete pulses on the screen of a cathode-ray oscilloscope. A grid of 1 cm squares covers the screen.

Question 18

The diagram shows two complete pulses on the screen of a cathode-ray oscilloscope. A grid of 1 cm squares covers the screen. The time-base setting is 1 μs cm^-1.

How long does each pulse last?

A 2 μs              B 3 μs              C 4 μs              D 6 μs

Reference: Past Exam Paper – June 2012 Paper 12 Q6

Solution:

Answer: A.

Horizontally, the divisions gives the time while vertically, the division represents the voltage.

Each pulse occupies 2 cm horizontally.

Time-base setting = 1 μs cm^-1

Duration of pulse = 2 × 1 = 2 μs

The diagram shows the waveform of a signal displayed on a cathode-ray oscilloscope.

Question 15

The diagram shows the waveform of a signal displayed on a cathode-ray oscilloscope.

The time-base is set at 5.0 ms per division.

The Y-gain is set at 5.0 mV per division.

What are the amplitude and the frequency of the signal?

amplitude / mV            frequency / Hz

A                     10                                50

B                     10                                100

C                     20                                50

D                     20                                100

Reference: Past Exam Paper – November 2018 Paper 11 Q23

Solution:

Answer: A.

The amplitude is measured (vertically) starting from the central line to the peak.

Y-gain = 5.0 mV per div

Amplitude = 2 div = 2 × 5.0 = 10 mV

The period is measured horizontally. One complete wave occupies 4 div.

Time-base setting = 5.0 ms per div

Period T = 4 div = 4 × 5.0 = 20 ms = 20×10^-3 s

 

Frequency = 1 / T = 1 / (20×10^-3) = 50 Hz