[Music] hello welcome back to my channel today we will start a new chapter which is chapter nine simple harmony motion exercise number one the expressions of the displacement x of an object undergo a linear shm is given by x equals to 3.5 sine 8 pi t plus 0.25 pi okay so where x and t are measured in cm and also in second respectively okay determine the amplitude so if you compare with the equations where as a general x equals to a sine omega t plus v meaning that our amplitude a is actually 3.5 cm and we want to find the period where we knows that omega is equal to 8 pi and omega equals to 2 pi over t so therefore our period is equal to 2 pi over pi period that we will get is equal to 0.25 seconds next frequency period equals 1 over f so f will equals to 1 over period or 1 over 0.25 so therefore we will get 4 hertz b find the phase angle when t equals to 0.02 so we know that the phase angle here we are referring to this one eight pi t plus zero point two five pi okay so we substitute our phase angle is two pi t plus zero point two five pi so we substitute t equals to 0.02 okay so 8 pi 0.02 plus 0.25 pi okay so if we press calculator we will get 0.41 pi in radian okay so this is our phase angle see find the displacement displacement here we are referring to the x and the instance mentioned above so from the equation x equals 3.5 sine 8 pi t so we substitute t equals to 0.02 okay plus 0.25 okay so if you press calculator we will get 3.4 cm so remember the value here is in radian okay so remember you must change it become radiant in the mode you must change mode calculator mode in radian okay because this one is no longer in degree so remember to change calculator in radian okay mode change to radiant questions number two one is the equation describing the motions of the mass on the end of the vertical spring which is stretch 8.8 cm from the equilibrium position and then we release from rest okay and the period is 0.66 seconds so what will be its displacement after 1.8 seconds okay so meaning that this is the vertical spring where it is stretched downward okay so for one complete cycle it will from amplitude here it will go up to the equilibrium position and then it will go up to then positive amplitude after that it will go down to equilibrium position and finally it will go back to its original position the graph given generally we can write x equals to negative a cos omega t okay so we substitute all the informations that we have where the spring is stretched 8.8 cm so this is actually the amplitude because this is the maximum displacement you can go and the period is equal to 0.66 seconds okay so we substitute the amplitude is equal to 8.8 cm and cos omega omega we know that is equals to 2 pi over t where t is equal to 0.66 so equations that we get is equal to negative 8.8 cos 3.03 pi t okay so this is the equations that we get okay why the equation is different with the equations given because we write in negative cost okay if you write it in sign meaning that initially if you want to change negative cost into sign if you want to write sine equation meaning that our graph we should start from here okay but initially the questions give us our graph is start from the negative and precue okay so meaning that our y exit okay the y and z we must shift to the left okay when we shift to the left our equation become x equals to a sine omega t so if you shift to the left meaning that the phase difference here the phase difference here between the sine graph and our graph okay is equal to you must minus you must minus with pi over 2 okay or if you don't want to write in sine you also can write in x equals to negative a cos omega t also can okay so both also acceptable so if let's say you want to change it into sine you also can write it as x equals to a sine omega is 2 pi over 0.66 t minus pi over 2 okay so equations that we will get is equal to 8.8 sine 3.03 pi t minus pi over 2 okay so both also accept the book okay next one we want to find what is its displacement when uh t equals to 1.8 seconds so we substitute into either either cos or sine okay so if let's say i substitute into negative cos equation x will equal to negative 8.8 cos 3.03 pi n times 1.8 okay so finally we will get 2.72 because 8.8 in cm so our answer also in cm okay question number three a particle undergo simple harmonic motions along a straight line completing 20 cycle into it in two seconds okay so actually this is the information of omega where omega equals to 20 cycles so it's times two pi in two seconds okay so 2 and 2 we can cancel off the omega that we get is equal to 20 pi radian per second and the amplitude of the motion is in 3 cm so amplitude a here is equal to 3 cm okay so here when t equals to 0 the particle is at equilibrium position meaning that x is equal to a sine omega t because it starts from equilibrium position displacement to the right of o is considered to be positive so a write down an expression representing the displacement of the particle okay so we know because it's not from equilibrium so we will get a sine graph so we substitute into the equation a is 3cm omega is equal to 20 pi t okay so this is the equations for the displacement b determine the displacement of the particle from o when t goes to 1 over 6 so we substitute into the equation when t equals 1 over 16 so we substitute 3 sine 20 pi t where t we substitute 1 over 16 okay and remember all this value you must change it into a radian mode okay because this is not a degree so you must convert change the mode calculator mode into radian okay so after that if you press calculator we will get negative 2.1 cm because tier 3 is still in cm and next we continue with c determine the time when the particle axis equals to positive 1.5 cm so we substitute x equals to a sine 20 pi t where x equals to 1.5 a is three so both is cm and this one also cm so later we will cancel off okay sine 20 pi t 1.5 over 3 equals to sine 20 pi t outside 1.5 over 3 so we will get a sine 0.5 equals to 20 pi times t okay so finally t that we will get is equal to 8.33 times 10 to the power of negative 3 in second okay next we goes to questions number four where questions number four they give you a graph f versus the displacement a variable force f acting on an object of the mass okay so that gives you the mass 0.15 cause the object undergo linear simple harmony motion figure shows how the fronts vary with the displacement of the object for each cycle of the oscillation determine the amplitude and also we want to find the frequency so what is the relationship here is we know that f equals to m a and we also know that a just now a equals to negative omega square x so we substitute negative omega square times x when f equals to 3 when f equals to 3 our displacement is equal to negative 0.2 so we substitute into the equation 3 equals to m 0.15 omega we don't know x is equal to negative 0.2 okay so 3 over 0.15 negative negative become positive 0.2 omega squared so finally omega that we will get is equal to third 100 or is equal to 10 okay but we want to find the frequency the frequency so the frequency is equal to 2 pi f equals to 10 so therefore frequency is equal to 10 over 2 pi grade c that we will get here is equal to 1.6 hertz next we go through and pre cube so if you refer here for x axis here we have negative 0.2 and on the right hand side we have 0.02 to find the amplitude amplitude is equals to the maximum displacement so the amplitude equals to x maximum so therefore we will take 0.2 because 0.2 is greater than 0.02 okay so we will take 0.2 meter because here the unit in meter okay so that's all for today thank you class see you on next video bye [Music] you