Respuesta :
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = [tex]\sqrt{ \frac{T}{ \mu } }[/tex]
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ = [tex]\sqrt{ \frac{T_o}{ \mu_o} }[/tex]
with the new dough
v = [tex]\sqrt{ \frac{T_o}{ 2 \mu_o} }[/tex]
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v = [tex]\sqrt{ \frac{T_o}{ 3 \mu_o} }[/tex]
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v = [tex]\sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }[/tex]
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v = [tex]\sqrt{ \frac{T_o}{5 \mu_o} }[/tex]
v = 1 /√5 v₀
v = 0.447 v₀
(a) When the mass is doubled, the speed of the wave in the string is 0.7071 v₀.
(b) When the wire is replaced with an identical one, the speed of the wave will be the same.
(c) When the mass of the wire is tripled, the speed of the wave in the string is 0.577 v₀.
(d) When the length is doubled, the speed of the wave in the string is 1.414 v₀.
(e) When the mass increases by a factor of 10 and the length doubles, the speed of the wave in the string is 0.447 v₀.
The given parameters:
- mass, M
- Length, L
- Speed of the wave, = V
- Tension on the string, = T
The speed of the wave at a given tension and the mass per unit length;
[tex]v_0 = \sqrt{\frac{T}{\mu} } \\\\v _0= \sqrt{\frac{T}{M/L} } \\\\v_0 = \sqrt{\frac{LT}{M} }[/tex]
(a) When the mass is doubled,
[tex]v = \sqrt{\frac{LT}{2M} } \\\\v = \sqrt{\frac{1}{2} } \times \sqrt{\frac{LT}{M} }\\\\v = \sqrt{\frac{1}{2} } \times v_0\\\\v = 0.7071 v_0[/tex]
(b) When the wire is replaced with an identical one,
the mass per unit length will be constant and the speed will be the same
v = v₀
(c) When the mass of the wire is tripled;
[tex]v = \sqrt{\frac{LT}{3M} } \\\\v = \sqrt{\frac{1}{3} } \times \sqrt{\frac{LT}{M} } \\\\v = \sqrt{\frac{1}{3} } \times v_0\\\\v = 0.577 \ v_0[/tex]
(d) When the length is doubled;
[tex]v = \sqrt{\frac{2LT}{M} } \\\\v = \sqrt{2} \times \sqrt{\frac{LT}{M} } \\\\v = \sqrt{2} \times v_0\\\\v= 1.414 v_0[/tex]
(e) When the mass increases by a factor of 10 and the length doubles;
[tex]v = \sqrt{\frac{2LT}{10M} } \\\\v = \sqrt{\frac{1}{5} } \times \sqrt{\frac{LT}{M} }\\\\v = \sqrt{\frac{1}{5} } \times v_0\\\\v = 0.447 \ v_0[/tex]
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