21.3 Transverse Standing Waves 641 If there are no nodes between the two at 1.0

Question

21.3 Transverse Standing Waves 641 If there are no nodes between the two at 1.0 m and 1.5 m, EXAMPLE 21.2 A standing wave on a string A 2.50-m-long string vibrates as a 100 Hz standing wave with then the node spacing is A/2 -0.50 m. The number of 0.50-m- nodes 1.00 m and 1.50 m from one end of the string and at no wide segments that fit into a 2.50 m length is five, so this is the points in between these two. Which harmonic is this, and what is m 5 mode and 100 Hz is the fifth harmonic. The harmonic fre- the string's fundamental frequency MODEL The nodes of a standing wave are spaced /2 apart. VISUALIZE The standing wave looks like Figure 21.5 SOLVE quencies are ; hence the fundamental frequency is 100HZ-20Hz -Is- STOP TO THINK 31.3 A standing wave on a string vibrates as shown at the right. Suppose the string tension is quadrupled while the frequency and the length of the string are held constant, which standing-wave pattern is produced? Original standing wave Standing Electromagnetic Waves

Explanation / Answer

wave velocity = sqrt[ T / mu ]

if T' = 4 T

then v' = sqrt(4) v = 2 v  

and wavelength = v / f

wavelength' = 2 wavelegth

for given case, wavelength = L

new wavelength = 2 L

hence it will be in fundamental node.

Ans(a) { in this case, wavelength = 2L }

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