Ask question

Ask question

All categories

2 years ago

12

One of the harmonics of a string fixed at both ends has a frequency of 52.2 Hz and the next higher harmonic has a frequency of 6

0.9 Hz. What is the fundamental frequency of the string?

1 answer:

Angelina_Jolie [31]2 years ago

6 0

Solution :

Frequency may be defined as the number of observation or number of waves that is taken in per unit time. The unit of frequency is Hertz or Hz.

It is given that :

Successive harmonic frequencies, f = 52.2 Hz

and f' = 60.9 Hz

Therefore, fundamental frequency, F = f' - f

F = 60.9 - 52.2

F = 8.7 Hz

Therefore the string which is fixed at both the ends forms all the harmonics.

You might be interested in

A girl is out jogging at 2.00 m/s and accelerates at 1.50 m/s^2 until she reaches a velocity of 5.00 m/s. How far does she get?

Maksim231197 [3]

Answer:

7.00 m

Explanation:

Given:

v₀ = 2.00 m/s

v = 5.00 m/s

a = 1.50 m/s²

Find: Δx

v² = v₀² + 2aΔx

(5.00 m/s)² = (2.00 m/s)² + 2(1.50 m/s²)Δx

Δx = 7.00 m

5 0

2 years ago

jarptica [38.1K]

Answer:

5.03

Explanation:

trust me

7 0

2 years ago

Which statement best describes how the temperature of the oceans surface water varies

lorasvet [3.4K]

If the temperature is high there is less water because it evaporates if it is cloudy it is more because it doesn't evaporate

8 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

2 years ago