A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to f
1 answer:
<h2>The frequency of driver is 700 Hz</h2>
Explanation:
The frequency of wave in a string is given by the relation
n =
here n is the frequency
p is the number of antinodes and l is the length of string .
T is the tension in string and m is the mass per unit length
Thus 420 = I
Now if there is 5 antinodes , the value of p = 5
Thus n = II
Dividing II by I , we have
n/420 = 5/3
or n = 5/3 x 420 = 700 Hz
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Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
