Question

3. A 250 g mass is attached to a horizontal spring and oscillates with a frequency of 2.1 Hz. At one instant the mass is at -4.3 cm and has a horizontal velocity of 25 cm/s.

A. What is the spring constant?

B. What is the total energy of the oscillator?

C. What is the period of oscillation?

D. What is the amplitude?

E. What is the maximum speed?

Answer 1

Understanding the whole concept and then solving the problem

=> under standing the concept. Ka spring /foru constant. Mặc tra Vg=m/s Hmmm 2=-20 2-0 x = 20 mean positron. As we can see that 3 positions of spring is shown above O are the catremifres 20= amplitude Now, if the string is streched to any position for a value = type of there are two Alternate Energies Energies stored in then as we know that mass - the spring Us 1) Potential Energy it occurs due to chonge in position of serving (m) Ves = { Kz? K =) 2) Kinetic Energy - it occurs due to veloudy of mawimi Kas & more so, total energy of system (neglecking all other factors are L Ę Us + Kx | at any arbitrary displacement of a و جا

• Now, studying at position o since its the extreeme position on left side so velo by at this point would be V-om/s aus the direchon al velou'ly would be chonging at this position Velouty offer velouty (Vom/s) position Before also the cusplaument would be a 2 -20 so, total energy of systm at this posrben would be Exo U.20+ Kogo *(-20) + 4 m(je ' Exo = 1 *x* 1 o. virom/s I ka o now, studying at position equilibrium /mean position ! as we know that at mean hosshion Velocity is maximum => V2 - Vo = Vman & x=0 ) no displa ament so, tofal Energy at position @ would be Es: Ub + Ks = 1 K603* +4 ()' = { mm fo most @

Now, Studying at positon ( X=20 Vasomys since, its extreme position and velocity would also be changing its direction. 00 would be So, tol al energy at this position Exo = Uxo + Kxo = KC20? + + mvg's į Kroč + 0 1 Eao= 1 K6² | conserved always ( so, as per law of conservation of energy, Energy should be son form equa trono 0, 0, m s Exz Eoz Entos Exo a total Energy of system so, at any pornt the value of total Energy of system could be Et £ Kao? = {mvo' = { Kap+LmVA? always conserved.

Now using the above concepts in our problem

y = Now , Solving the given problem (3) geven, m= 250g = 0.250 Kg frequency - f= 2.1 Hz at a particular instant s position = x= -4.3 cm = -0.043 m velocity = Vote 25 cm/s = 0.25m/s Assuming that the position was taken from the mean position of spoing (taking mean position as origin) for con I compani mean post'hon

bane , Assuming the spring constant = K & angular frequencya W= 2nf = 2 x 2 = J 1 in 2012 Jan & squaxing both sides 41?$2= Km = K = mx 47°42= 0-250 x4xx (2-2) K = 43-52S N/m - 4. 3525 x10 N/m | part (B) * E Total Ency9y Oscillatoy at any instant = E E- Uz + Kx - kx + Im E = * (43-525) 60.00z) + { *(0-250) (0-252) F = 4.805 xro”] part (c) 19 per'od of oscillation - T= 21 T= 0.476 s = 4.76xló's part (0) > Assuming the amplitude = Xo so, from law of Conservation of Energy Es Ex= 4Kxé , 4.805x16?J xo=(2x4-805 x10² m V 43.525 20 - 4.70 x162 m = 0.047m - Amplitude

þarrt (E) UO » Assuming maximum speed= Vo so, again from law of conservation of energy, E=E2 = 1 Kao = £m Vo² = 4-805 710² Vo = ? 41-305 162 m/s Vos 2 x 4.805x10 0.250 Į Vo= 62 X16'm/s = 0.62 m/s y masumum speed.