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Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
The given data is as follows.
Inner wheel Radius = 20 m,
Distance between left and right wheel = 1.5m,
Let us assume speed of drive shaft is N rpm.
Formula to calculate angular velocity is as follows.
Angular velocity of automobile = w =
where, V = linear velocity of automobile m/min,
R = turning radius from automobile center in meter
In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.
Now, we assume that
u = linear velocity of inner wheel
and, u' = linear velocity of outer wheel.
Formula for angular velocity of inner wheel w = ,
Formula for angular velocity of outer wheel w =
Now, for inner wheels
w =
=
u =
=
If radius of wheel is r it will cover distance in one min.
Since, velocity of wheel is u it will cover distance u in unit time(min)
Thus, u = =
Now, rotation per minute of inner wheel is calculated as follows.
n =
= (since 2d = 1.5m given, d = 0.75m),
=
So, rotation per minute of outer wheel; n' =
=
=