An object is lifted upward at a constant speed. The gain in potential energy of the object is due to the
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Electric field between the plates of parallel plate capacitor is given as
here area of plates of capacitor is given as
also the maximum field strength is given as
now we will plug in all data to find the maximum possible charge on capacitor plates
so the maximum charge that plate will hold will be given by above
Answer:
(a) Wavelength is 0.436 m
(b) Length is 0.872 m
(c) 11.518 m/s
Solution:
As per the question:
The eqn of the displacement is given by:
(1)
n = 4
Now,
We know the standard eqn is given by:
(2)
Now, on comparing eqn (1) and (2):
A = 1.22 cm
K =
where
A = Amplitude
K = Propagation constant
= angular velocity
Now, to calculate the string's wavelength,
(a)
where
K = propagation vector
(b) The length of the string is given by:
(c) Now, we first find the frequency of the wave:
Now,
Speed of the wave is given by: