A mass of 050 kg hangs from a light spring and executes shm...

A mass of 0.50 kg hangs from a light spring and executes SHM so that its position x is given by x = A cos ωt. It is found that the mass completes 20 cycles of oscillation in 80 s. (a) Determine (i) the period of the oscillations, (ii) the angular frequency of the oscillations and (iii) the spring constant k. (b) Using a value of A = 2 mm, make sketches of the variations with time t of the displacement, velocity and acceleration of the mass.

Transcript

00:01 In this problem we have been given that there is a mass which is hung with the help of light inextensible string here.

00:09 And this block is having mass of 0 .5 kilograms.

00:14 And it is given that it is executing simple harmonic motion whose displacement is given as x equals to a times cos omega -t.

00:27 And in this oscillation, it is found that the mass completes 20 cycles in 80 seconds.

00:37 So 20 cycles is completed in 80 seconds.

00:41 That's what we have been given here.

00:43 So we need to determine first the time period.

00:50 So here we can observe and get the frequency.

00:54 So 20 cycle is completed in 80 seconds.

00:57 That means one cycle will be completed in 80 by 20 seconds.

01:04 So that would be four seconds.

01:07 So here the frequency comes out to be 1 by 4 hertz.

01:13 And using this frequency, we can get the time period by reciprocating this frequency.

01:19 So that comes out to be 4 seconds.

01:21 Now that's the time period of oscillation of this block.

01:25 And now we need to figure out the angular frequency.

01:29 So from the frequency that we have computed in part one, we can figure out the angular frequency omega by using 2 pi f.

01:37 So we multiply 2 pi with the frequency 1 by 4.

01:41 So that comes out to be 2 pi radiance per second.

01:45 So when we simplify, we get this as 6 .28 radiance per second.

01:51 Now this is the angular frequency of this mass.

01:58 And next we need to figure out.

02:00 The spring constant.

02:03 So in order to get the spring constant, we use the idea that k is equal to or k by m is equal to omega square.

02:15 So from here we can rearrange and get the spring constant by one multiplying mass with omega square.

02:22 So we take the mass that's 0 .5 and take the square of omega that we just now figured out...

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