A spring stretches by 2 mm when it is loaded with a mass of 200 g. From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maximum energy in the spring are Hz and J, respectively. (Take g = 10 m/s2)
- 5 50 and 8 10-3
- 5 50 and 8
- 10 50 and 2 10-3
- 5 50 and 16 10-3
Answer
(1) 5 50 and 8 10-3
Solution
Spring constant from the static stretch: k= mgx0= 0.2 102 10-3=1000 N/m
Frequency: f= 12 km= 12 10000.2= 50002
⇒ f= 10 502 = 5 50 Hz
The mass oscillates with amplitude 2 mm about the equilibrium position, so at the lowest point the spring is stretched by xmax=2+2=4 mm
∴ Umax= 12kxmax2= 12 1000 (4 10-3 )2
Umax=8 10-3 J
Answer: (1)