4 a mass weighing 3 lb stretches a spring 3 in if the mass...

4. A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in and then set in motion with a downward velocity of 2 ft/s, and if there is no damping, find the position u of the mass at any time t. Determine the frequency, period, amplitude, and phase of the motion. 6. A spring is stretched 10 cm by a force of 3 N. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 N when the velocity of the mass is 5 m/s. If the mass is pulled down 5 cm below its equilibrium position and given an initial downward velocity of 10 cm/s, determine its position u at any time t. Find the quasi-frequency μ and the ratio of μ to the natural frequency of the corresponding undamped motion.

Transcript

00:01 Hi, here for the given question we need to find the value of position of a u of a mass at any time.

00:10 Here there is no damping so the equation which we are going to use is m multiplied with u double dash of t plus k u of t is equal to zero.

00:23 Now here in this question we are given that mass m equals to 3 lb which can be written as 1 .36 kg.

00:32 Further force f is equal to 13 .34 newton and here the initial displacement in the string is x is equal to 3 inches which can be also written as 0 .0762 meters.

00:55 So here k is equal to force upon displacement so this is again equal to 13 .34 divided by 0 .0762.

01:05 So calculating this here we have 174 .95 newton per meter.

01:12 Now here further our equation can be written as 1 .36 u dash of t u double dash of t plus 174 .95 u of t is equal to zero.

01:28 Now here in order to solve this equation we will take the value of u of t.

01:34 So here we know that u of t is equal to a cos omega t plus now here in our case double derivative can be written as u double dash of t equals to minus a omega square cos of omega t plus.

01:53 So here now substituting this value in our above equation let this be our equation number one.

02:00 In equation number one here we have minus 1 .36 a multiplied with omega square multiplied with cos of omega t plus 5 plus 174 .95 multiplied with a multiplied with cos of omega t plus 5 is equal to 0.

02:19 Now here as we can see that if we compare the coefficient then here we can observe that the value of 1 .36 omega square minus plus 174 .95 is equal to 0.

02:42 So here on solving this we have value of omega equals to 11 .34.

02:49 Now here using omega we can find the value of frequency.

02:53 So frequency f is equal to omega upon 2 pi.

02:56 So here we have 11 .34 divided by 2 times 3 .14.

03:01 So calculating this we have frequency equals to 1 .805 hertz and here the time t equals to 1 upon f.

03:09 So calculating this we have time period is equal to 0 .554 seconds.

03:16 Now here we need to calculate the maximum amplitude.

03:19 So here the maximum amplitude is the maximum distance.

03:23 So here the maximum distance is 0 .0762 meter.

03:29 This is the initial stretch.

03:30 So here now this is our required basic condition we needed in order to calculate the value of u of t.

03:39 So here in our case we can say that u of t is equal to 0 .0762 multiplied with cos of 11 .34 t plus 5.

03:49 So this is our first part.

03:51 Now here further moving towards the second part of the question.

03:55 Here in the second part we need to calculate what is the value of u of t only.

04:02 So here the equation we have m multiplied with u double dash of t plus c multiplied with u dash of t plus k multiplied with u of t is equal to 0...

Question

Please give Ace some feedback