00:01
Hi, in this question we have this diagram in which we have a steel wire having length l given as 2 meter and mass of the steel wire is 35 grams, that is 35 into 10 raised to power minus 3 k.
00:15
So it is used to stretch a spring having spring constant k 179 newton per meter by hanging a mass.
00:24
So this is the mass of the wire and this mass small m is given as 9 .4.
00:30
Kg.
00:32
Now, when the system is in equilibrium, the extension of the spring, so the extent up to which it gets extended is given as x equals to 0 .514 meter.
00:47
Exactly one half of the wire is between the pulley and the spring and the upper half of the wire oscillates with the pattern shown.
01:00
So in part, we have to calculate the frequency f of the wires oscillation.
01:06
Now we know that speed of wire v is equals to under root t by new.
01:14
So, v is equals to under root x divided by m by l.
01:22
And the frequency, that is f, is given as v by lambda.
01:29
So on placing the value of v, we can write 1 by lambda under root kxl by m.
01:39
So frequency f will be.
01:42
So since length is 2, so wavelength is half of the length.
01:46
So this is 1 by 1 under root 179 into 0 .514 into 2 divided by 35 into 10 raise 2...