One of the 63.5-cm-long strings of an ordinaryguitar is tuned to produce the note {\rm B_3}(frequency 245 Hz) when vibrating ini
1 answer:
Answer:
a,V=311.15m/s
b.246Hz
c.245Hz
d. 1.4m
Explanation:
One of the 63.5-cm-long strings of an ordinaryguitar is tuned to produce the note {\rm B_3}(frequency 245 Hz) when vibrating inits fundamental mode.
A.Find the speed of transverse waves on this string.B.If the tension in this string is increased by1.0%, what will be the new fundamental frequency ofthe string?C.If the speed of sound in the surrounding air is 344m/s, find the frequency of the sound wave produced in theair by the vibration of the {\rm B_3} string.D.If the speed of sound in the surrounding air is 344m/s, find the wavelength of the sound wave produced in theair by the vibration of the {\rm B_3} string
NOTE
speed is distance by the wave per time
frequency is the number of oscillation the wave front makes in one seconds
the wave speed is given by
recall also that the wave speed is v=f lambda
for a standing wave , we know the fundamental frequency of a string is
f1=v/2L
L=length of the string
f1=245Hz
V=?
L=0.635m
V=245*2*0.635
V=311.15m/s
b. tension in the string is increased by 1%
F2=F+1%
f2=101F%
substituting for F2 into this equation
v2=
v2=1.01^0.5*311.15m/s
v2=312.7m/s
for the new fundamental frequency we have
f2=312.7/2*0.635
f2=246Hz
c. the frequency of the sound wave equal the frequency of the string that created it
c. fs=245Hz
d. speed of sound in air344m/s
v=344m/s
v=f*lambda
lambda is the wavelength
344=245*lambda
344/245=1.40m
wavelength of string B3 is 1.4m
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Answer:
1) Hence, the period is 0.33 s.
2) The amplitude is 10 cm.
Explanation:
1) The period is given by:
Where:
f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz
Hence, the period is 0.33 s.
2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:
Therefore, the amplitude is 10 cm.
I hope it helps you!
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s