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1750 N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (A and B), each 1.25 m long and weighing 0.290 N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached. If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling

Sagot :

The time delay between the arrival of the two pulses at the ceiling is 2.33 ms.

What is center of gravity?

Any object of mass is attracted towards another by point where whole mass is supposed to be concentrated is called center of gravity.

A 1750 N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (A and B), each 1.25 m long and weighing 0.290 N.

The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached.

Taking moments about one end, to find the tension at other end.

Tb X L = WxL/3

here, W is the weight, L is the length of wire and Tb is the tension at point B.

So, Tb = W/3

and
Tb - Ta = W

here, Ta is the tension in end A.

Substitute the values, we get

Tb = 583.33N and Ta = 1166.66N

Velocity of end A, Va = (1166.66 x1.25 x 9.81/0.290 )^(1/2)

Va = 222 m/s

Velocity of end B, Vb = (583.33 x1.25 x 9.81/0.290 )^(1/2)

Va = 157 m/s

Time delay t is related with the velocities as

t = L/Vb - L/Va

Substitute the values, we get

t= 0.00233 s

t = 2.33 ms

Thus, time delay between the arrival of the two pulses at the ceiling is 2.33 ms.

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