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At
t = 0,
the instantaneous position of two pulses moving along a taut
string with a speed
v = 2.67 cm/s
are as shown in the diagram below. Each unit on the
horizontal axis is 2.0 cm and each unit on the vertical axis is
4.0 cm.
(a) At what location will the resultant of the two pulses have
minimum amplitude?
____ cm
(b) At what time will the resultant of the two pulses have
minimum amplitude?
____ s
(c) What is the value of this minimum amplitude?
____ cm pulse 1 Pulse 2
Solution
Horizontal distance between pulses S = 10(2 cm) = 20 cm
Relative velocity V = v -(-v) = v+v = 2v = 2(2.67 cm/s) = 5.34
cm/s
(a) At what location will the resultant of the two pulses have
minimum amplitude = center between two pulses
= 7 cm
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(b) At what time will the resultant of the two pulses have
minimum amplitude t = S / V
t = 20 cm / 5.34 cm/s
= 3.745 s
(c) the value of this minimum amplitude = 4(4cm) -3(4cm) =
16 cm - 12 cm = 4 cm

### Unformatted Attachment Preview

At t = 0, the instantaneous position of two pulses moving along a taut string with a speed v = 2.67 cm/s are as shown in the diagram below. Each unit on the horizontal axis is 2.0 cm and each unit on the vertical axis is 4.0 cm. (a) At what location will the resultant of the two pulses have minimum amplitude? ____ cm (b) At what time will the resultant of the two pulses have minimum amplitude? ____ s (c) What is the value of this minimum amplitude? ____ cm pulse 1 Pulse 2 Solution Horizontal distance between pulses S = 10(2 cm) = 20 cm Relative velocity V = v -(-v) = v+v = 2v = 2(2.67 cm/s) = 5.34 cm/s (a) At what location will the resultant of the two pulses have minimum amplitude = center between two pulses = 7 cm (b) At what time will the resultant of the two pulses have minimum amplitude t = S / V t = 20 cm / 5.34 cm/s = 3.745 s (c) the value of this minimum amplitude = 4(4cm) -3(4cm) = 16 cm - 12 cm = 4 cm Name: Description: ...
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