1- Two strings which are fixed at both ends are identical except that one is 0.6
ID: 1644874 • Letter: 1
Question
1- Two strings which are fixed at both ends are identical except that one is 0.62 cm longer than the other. Waves on both of these string propagate with a speed of 34.8 m/sec and the fundamental frequency of the shorter string is 229 Hz.
a- What is frequency of the beat that would result if these two strings were plucked at the same time?
b- What is the beat frequency if the length difference is now 0.77 cm?
2- A physics student sits by the open window on a train moving at 25 m/sec towards the east. Her boyfriend is standing on the station platform, sadly watching her leave. When the train is 100 meters form the station, it emits a whistle at a frequency of 4 × 103 Hz.
a- A chemistry student (with nothing better to do than ride around on trains all day) is on a west bound train moving at a velocity of 15 m/sec towards the station. Before the two trains pass each other, what frequency will he hear from the train whistle from the physics student's train?
b- What frequency will the chemistry student hear after his train passes the physics student's train?
Explanation / Answer
Solution:
1.
As we know;
f1 =v/2L = 229
=> Lshort = 34.8 / (2*229)
=> Lshort = 0.0759825 m or 7.6 cm = L1
=> LLonger = 8.22 cm L2
Hence;
f1 = v/2L1 and f2 = 2*(v/2L2)
=> f1 = 229 Hz and f2 = 211.68 Hz
and as we know;
Fbeat= absolute value ( f 1- f 2 )
=>Fbeat = 17.32 Hz
b)
Similarly:
LLonger = 8.37 cm = L2
Hence;
f1 = v/2L1 and f2 = 2*(v/2L2)
=> f1 = 229 Hz and f2 = 415.77 Hz
and as we know;
Fbeat= absolute value ( f 1- f 2 )
=>Fbeat = 415-229 Hz
=>Fbeat = 186.77 Hz
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