Two lab partners, Mary and Paul, are both farsighted. Mary has a near point of 68.6 cm from her eyes and Paul has a near point o
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Answer:
(a) the fundamental frequency of this string is 65 Hz
(b) the harmonics of the given frequencies are third and fourth respectively.
(c) the length of the string is 2.74 m
Explanation:
Given;
mass density of the string, μ = 3 x 10⁻³ kg/m
tension of the string, T = 380 N
resonating frequencies, 195 Hz and 260 N
For the given resonant frequencies;
(c) From any of the equations, solve for Length of the string (L);
(a) the fundamental frequency is calculated as;
(b) harmonics of the given frequencies;
the first harmonic (n = 1) = f₀ = 65 Hz
the second harmonic (n = 2) = 2f₀ = 130 Hz
the third harmonic (n = 3) = 3f₀ = 195 Hz
the fourth harmonic (n = 4) = 4f₀ = 260 Hz
Thus, the harmonics of the given frequencies are third and fourth respectively.
Explanation:
It is given that,
length of steel wire, l = 0.75 m
Mass of the wire, m = 12 g = 0.012 kg
Fundamental frequency, f = 120 Hz
We need to find the mass of the anvil (m'). The fundamental frequency is given by :
v is the speed of the mass
Speed is given by :
is the mass per unit length,
T is the tension in the wire,
T = 518.4 N
Tension in the wire, T = m' g
m' = 52.89 kg
So, the mass of the anvil is 52.89 kg. Hence, this is the required solution.