00:01
Hello guys, so for the given question we have been provided that the initial mass attached to the spring m1 is 0 .50 kilogram with which the spring is oscillating with the time period of t1 is equal to 1 .5 seconds.
00:16
Now if the time period of oscillation of the spring becomes t2 equals to two seconds, we need to calculate the value of mass attached to the spring for this situation.
00:26
So for the solution of this problem, we first need to understand a formula for time period of oscillation of a spring, which is given by t is equals to 2 pi multiplied with under root of m divided by k.
00:45
Here t is the time period, m is the mass and k is the spring constant.
01:02
All right so for case one we can write the time period of oscillation t1 is equal to 2 pi multiplied with m1 divided by k placed in the square root all right or 1 .5 will be equal to 2 pi divided by under root of m1 divided by k from here the value of k, that is the spring constant, comes out to be equal to 8 .77 newton per meter.
01:56
Now, for case 2, we can write time period of oscillation t2 will be equals to 2 pi multiplied with the square root of m2 divided by k...