he position of a mass-on-a-spring oscillator is given by
1.
a) Kinetic energy is equal to 1/2*M*V^2. Here m
(mass) is known which is 1.7Kg. now velocity of SHM or spring
oscillator is w.sqrt(A^2-x^2) where w= angular frequency=15,
A=amplitude=.34 so velocity will be max when displacement will be
zero so max velocity = w.A=15*.34=5.1. So max
K.E. = 1/2 * M*V^2 = 1/2*2.2*5.1*5.1 = 28.611
J.
b) K.E. = 1/2 * M*V^2=K.E. = 1/2 *
M*(A*w)^2
Now K.E. is proportional to square of Amplitude,
hence when K.E. becom twice, Amplitude will be sqrt(2) times of
A.
2.Y = stress/strain = F/A/(?L/L) = F*L/(A*?L) = 146N*0.69m/(?*(0.6x10^-3m)^2 x0. 41x10^-3m)
= 21.7x10^10Pa
3.y = 2.2 sin(23t)
You need to think of y = Asin(wt) type of equations where A is the
amplitude and w is the angular frequency of the
oscillation.
The max value of a sine function is +1 so the max value of y is
going to be 2.2 x 1 = 2.2 and this is the
Amplitude.
The min value of a sine function is -1 so the min value of y is
going to be 2.2 x -1 = -2.2 and this is the smallest value for
y.
You see that the wave oscillates between -2.2 and
+2.2
Notice that w = 23 so that the time period is 2pi/w = 2pi/23 ~
0.273 seconds.
the frequency is given by 1/T = 1/0.273 ~ 3.663 Hz
Hope this helps.
he position of a mass-on-a-spring oscillator is given by y = A sin(15t), where the value...
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