When an object is seen moving in relation to a stationary object, the stationary object is called the frame of reference or refe
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Answer:
The function is missing in the question. The function of the transverse pulse in the wire is given by
Explanation:
A transverse wave can be defined as the wave whose direction of displacement is always perpendicular to the direction of propagation. For example, surface wave at water bodies. While a pulse can be defined as a sudden change in a constant quantity such as a pulse of the radiation or current.
Let the wire of infinite length in both the directions and also the magnitude of deflection of wire be in the same shape except the point of maximum deflection to move along the wire.
Thus the equation of the pulse moving the in the positive x-direction moving at the speed of 2.10 m/s is
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Answer:
1) d = h- L - mg / k , 2) k = 2 mg/h (-1 +2 L / h)
Explanation:
1) Let's use the translational equilibrium equation
-W = 0
F_{e} = W
k x = mg
x = mg / k
from the statement of the exercise the height of the bridge, for the reference system in the river, is
h = L + x + d
x = h - L - d
we substitute
h - L -d = mg / k
d = h- L - mg / k
2) They ask us for the spring constant. For this part we can use energy conservation
Starting point. At the point before jumping
Em₀ = U = m g h
Final Point. When it's hanging
Em_f = + U = ½ k x² + mg d
Em₀ = Em_f
mg h = 1 / k x² + m g d
in the exercise they indicate that Kate touches the surface of the river, so the distance d = 0
mg h = ½ k x²
k = 2 mg h / x²
we substitute the value of x
k = 2mg h / (h -L)²
k = 2mg h / (- L + h)²
we simplify the expression
k = 2mg h / [h² (1- L / h)²]
k = 2m /h (1- L / h)⁻²
In these jumps the bridge height is always greater than the length of the rope L / h <1, so we can expand the last expression
(1- L / h)⁻² = 1 - 2 (1 -L / h) + 2 3/2! (1 -L / h)² + ...
for simplicity let's keep up to the linear term, we substitute in the solution
k = 2 mg/h [1 - 2 (1- L / h)]
k = 2 mg/h (-1 +2 L / h)