Effective frequency -- waves

  • Thread starter hidemi
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Homework Statement:
Two waves are described by y1 = 6 cos 180 t and y2 = 6 cos 186 t (both in meters). What effective frequency does the resultant vibration have at a point?
92 Hz B)183 Hz. C)6 Hz. D)3 Hz. E)366 Hz
The correct answer is B
Relevant Equations:
Y=−2Acos(At)cos(2πft)
Explanation:

y1=6cos(180t)
+) y2=6cos(186t)
y1+y2 =6cos(180t)+6cos(186t)y
=6(cos(180t)+cos(186t))

cos(A)+cos(B)=−2cos((A+B)/2)cos((A−B)/2)
y=6(−2cos((180+186)t/2)cos((180−186)t/2))
y=−12cos(183t)cos(3t)
y=(−12cos(3t))cos(183t)

Comparing with the standard equation,
Y=−2Acos(At)cos(2πft)

Therefore we have by comparison that,
2πf=183f=1832π
∴f=29.1Hz.


This is my answer. I could not find a matching answer. Where am I doing wrong?
 

Answers and Replies

  • #2
haruspex
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Yes, the question setter seems to have got confused between angular frequency and cycle frequency.
 

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