Question

There is a 3‑kg molecule is joined to an elastic band and experiences simple harmonic movement. In light of the data introduced in the chart and taking x0 = 4.4 cm and t1 = 0.58 sec, discover the spring constant of the elastic band.

X AM OV 2 - Xo

k = Unit in N/m

The molecule arrives at its maximum speed when passes the equilibrium, for instance, on occasion t1 and t2. What is this maximum speed?

v_max = Unit in cm/s

The 1st time the speed of the molecule is zero, when?

t = Unit in s

Answer 1

sol Given 3 kg Po = 4.4 cm ti 0.58 s where have lo Sinot This is the required displacement equation. will be maximum From Displacement Equation EL o.se cohen t O, 295 2 At t = 0.29 $ j x = so = 4.4 cm 4.4=4.4 SinWx0.29 Sin wxo.29 to X O 2 = 7/2=sines 3.14 5.41 rad 2x0: 29 S Angular frequency w 5.41 rad have and VIE force Constant Spring Constant 2 K = (5.41) K = 878 Now we have velocity Equation v=% Coswt how a sinut = v=dx dt at Bw Coswt) Vmax = ew 4.4 x 5.41 = 23.8 Cm cm Imax Speed Zero of the 23. 8 come molecule Zero when displacement at t = ti/2= 0.295 first time velocity 0.295 ie At t will be zero. Shot on V2