Someone please help!
2 answers:
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Answer:
(a) 0.42 m
(b) 20.16 N/m
(c) - 0.42 m
(d) - 0.21 m
(e) 17.3 s
Solution:
As per the question:
Mass, m = 0.56 kg
x(t) = (0.42 m)cos[cos(6 rad/s)t]
Now,
The general eqn is:
where
A = Amplitude
= angular frequency
t = time
Now, on comparing the given eqn with the general eqn:
(a) The amplitude of oscillation:
A = 0.42 m
(b) Spring constant k is given by:
Thus
(c) Position after one half period:
(d) After one third of the period:
(e) Time taken to get at x = - 0.10 m:
t = 17.3 s
Answer:
a) The closest object Anne can see, while wearing Bill's glasses is at approximately 22.39 cm relative to her eyes
b) The closest object Bill can see while wearing Anne's glasses is at approximately 26.38 cm relative to his eyes
Explanation:
The point where Bill has a near point = 125 cm
The point where Anne has a near point = 75.0 cm
Their vision are both corrected to the normal near point = 25.0 cm
The distance of their glasses from the eye, d = 2.0 cm
The lens formula is presented as follows;
The required distance of the of the object from Bill's glass, = 25 cm - 2.0 cm = 23.0 cm
The distance of the of the image from Bill's glass, = -(125 cm - 2.0 cm) = -123 cm
Therefore, for Bill, we have;
Plugging in the values gives;
Therefore;
= 2,829/100 cm = 28.29 cm
The required distance of the of the object from Anne's glass, = 25 cm - 2.0 cm = 23.0 cm
The distance of the of the image from Anne's glass, = -(75 cm - 2 cm) = -73 cm
The focal length for Anne is therefore;
Plugging in the values gives;
Therefore, we have;
= 1,679/50 cm = 33.58 cm
a) When Anne wears Bill's lasses, we have;
Therefore, we get;
≈ 20.39 cm
The distance of the closest object Anne can see, from her eye, =
+ d
∴ ≈ 20.39 cm + 2.0 cm = 22.39 cm
The distance of the closest object Anne can see, from her eye, ≈ 22.39 cm
b) The closest object that can be seen when Bill wears Anne's glasses, we have;
Therefore;
∴ ≈ 26.38 cm
The closest object Bill can see while wearing Anne's glasses, ≈ 26.38 cm.