please solve these descriptevly.... Since the time of ancient Rome, commanders o

Question

please solve these descriptevly....

Since the time of ancient Rome, commanders of armies have known that it is prudent to have

the troops break stride in their march when crossing a wooden bridge. This is because if the

troops march in a synchronized cadence, they produce a periodic coordinated jolt to the

bridge, which could excite one of its natural harmonic frequencies, causing a standing wave

to develop in the bridge.

a. If a marching army does create a standing wave in the bridge, what aspect of this

standing wave (A, f, T, ?) would be directly responsible for causing the bridge to break

apart? Explain.

b. Suppose a platoon comes upon a wood & rope bridge that is supported only at its two

ends. The commander stops his company short of the bridge and shakes the nearest end

of the bridge, testing to see if it seems strong enough to hold his men. The bridge ripples

all the way down its length, with the pulse re?ecting off the other side and returning, for a

roundtrip time of 2.5 seconds. Find the frequency of the fundamental harmonic standing wave for this bridge?

c. The commander decides the bridge is sturdy, and makes the tragic decision to order his

men to march on. Their marching pace is such that the frequency of footfalls matches the

5th harmonic frequency of the bridge. Sketch this standing wave, and calculate how

many steps they take per second.

Explanation / Answer

the main reason why this is done is resonance.if the natural frequency of the bridge matche with the frequency of the stride of the troop then the amplitude of vibration of the bridge will increase and be out of the safety limit and the bridge may even crumble.

so frequency is the important aspect to cinsider.

let L be the length of the bridge

in 1st harmonic

L = lamda / 2

v = f x lamda...................................1

v = velocity

lamda = wavelength

f = frequency

v = distance / time

v = 2L/2.5

so

from 1

2L/2.5 = f x 2L

f = 1/2.5

f = .4 Hz

fifth harmonic means there are 5 complete loops in the lenght L of the bridge

L = 5/2 x lamda

lamda = 2L/5

as the velocity will remain same

v = 2L/ 2.5 as in the previous case

but v = f x lamda

2L/2.5 = f x 2L/5

f = 5/2.5

f = 2 Hz

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