5 cm = 0.05 meters
T = period = 1/50 = 0.02 second
x = 0.05 sin 2 pi t/T = 0.05 sin 314 t
v = 0.05(314) cos 314 t
a = -0.05 (314)^2 sin 314 t
Now you should be able to figure it out
A body vibrate in s.h.m with a frequency of 50Hz and a mplitude of 4cm.find the period,the acceleration at the middle and at the end of the path oscillation,the velocity at the middle and at end of path of oscillation
5 answers
answer
That is not the Question
A body vibrate in SHM with a frequency of 5ohertz and amplitude of 4cm. Find:-
1. the period
2. the acceleration at the middle and at the end of the path of oscillation
3. the velocities at the middle and at the end of path of oscillation
4. the velocity and acceleration at a distance 2cm from a centre of oscillation
1. the period
2. the acceleration at the middle and at the end of the path of oscillation
3. the velocities at the middle and at the end of path of oscillation
4. the velocity and acceleration at a distance 2cm from a centre of oscillation
1.T=1/f =1/50=0.02s
2. The acceleration at the middle point is zero
ii). A(max)=Aw²
w=2πf
w²=4π²f²
A(max)=0.04×4×3.14²×50²
=3944m/s
2. The acceleration at the middle point is zero
ii). A(max)=Aw²
w=2πf
w²=4π²f²
A(max)=0.04×4×3.14²×50²
=3944m/s
3 answers