A standing wave is set up in a string of variable length and tension by a vibrator of variable frequency. Both ends of the string are fixed. When the vibrator has a frequency f , in a string of length L and under tension T , n antinodes are set up in the string. (a) If the length of the string is doubled, by what factor should the frequency be changed so that the same number of antinodes is produced? (b) If the frequency and length are held constant, what tension will produce n + 1 antinodes? (c) If the frequency is tripled and the length of the string is halved, by what factor should the tension he changed so that twice as many antinodes are produced?
A standing wave is set up in a string of variable length and tension by a vibrator of variable frequency. Both ends of the string are fixed. When the vibrator has a frequency f , in a string of length L and under tension T , n antinodes are set up in the string. (a) If the length of the string is doubled, by what factor should the frequency be changed so that the same number of antinodes is produced? (b) If the frequency and length are held constant, what tension will produce n + 1 antinodes? (c) If the frequency is tripled and the length of the string is halved, by what factor should the tension he changed so that twice as many antinodes are produced?
Solution Summary: The author explains that the frequency of the vibrator should be changed by a factor of 12 so that same number of 'antinodes' are produced.
A standing wave is set up in a string of variable length and tension by a vibrator of variable frequency. Both ends of the string are fixed. When the vibrator has a frequency f, in a string of length L and under tension T, n antinodes are set up in the string. (a) If the length of the string is doubled, by what factor should the frequency be changed so that the same number of antinodes is produced? (b) If the frequency and length are held constant, what tension will produce n + 1 antinodes? (c) If the frequency is tripled and the length of the string is halved, by what factor should the tension he changed so that twice as many antinodes are produced?
4.) The diagram shows the electric field lines of a positively charged conducting sphere of
radius R and charge Q.
A
B
Points A and B are located on the same field line.
A proton is placed at A and released from rest. The magnitude of the work done by the electric field in
moving the proton from A to B is 1.7×10-16 J. Point A is at a distance of 5.0×10-2m from the centre of
the sphere. Point B is at a distance of 1.0×10-1 m from the centre of the sphere.
(a) Explain why the electric potential decreases from A to B. [2]
(b) Draw, on the axes, the variation of electric potential V with distance r from the centre of the
sphere.
R
[2]
(c(i)) Calculate the electric potential difference between points A and B. [1]
(c(ii)) Determine the charge Q of the sphere. [2]
(d) The concept of potential is also used in the context of gravitational fields. Suggest why scientists
developed a common terminology to describe different types of fields. [1]
3.) The graph shows how current I varies with potential difference V across a component X.
904
80-
70-
60-
50-
I/MA
40-
30-
20-
10-
0+
0
0.5
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
VIV
Component X and a cell of negligible internal resistance are placed in a circuit.
A variable resistor R is connected in series with component X. The ammeter reads 20mA.
4.0V
4.0V
Component X and the cell are now placed in a potential divider circuit.
(a) Outline why component X is considered non-ohmic. [1]
(b(i)) Determine the resistance of the variable resistor. [3]
(b(ii)) Calculate the power dissipated in the circuit. [1]
(c(i)) State the range of current that the ammeter can measure as the slider S of the potential divider
is moved from Q to P. [1]
(c(ii)) Describe, by reference to your answer for (c)(i), the advantage of the potential divider
arrangement over the arrangement in (b).
1.) Two long parallel current-carrying wires P and Q are separated by 0.10 m. The current in wire P is 5.0 A.
The magnetic force on a length of 0.50 m of wire P due to the current in wire Q is 2.0 × 10-s N.
(a) State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P. [2]
(b) Calculate the current in wire Q. [2]
(c) Another current-carrying wire R is placed parallel to wires P and Q and halfway between them as shown.
wire P
wire R
wire Q
0.05 m
0.05 m
The net magnetic force on wire Q is now zero.
(c.i) State the direction of the current in R, relative to the current in P.[1]
(c.ii) Deduce the current in R. [2]
Chapter 18 Solutions
Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term
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