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REY [17]
3 years ago
10

A guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beat/s when sounded with a 357 Hz tuning fork. W

hat is the vibrational frequency (in Hz) of the string
Physics
1 answer:
STatiana [176]3 years ago
6 0

Answer:

349 Hz

Explanation:

We are told that the guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beat/s when sounded with a 357 Hz tuning fork.

This means that for the 352 Hz tuning fork, the vibrational frequency is;

f = 352 ± 3

f = (352 + 3) or (352 - 3)

f = 355 Hz or 349Hz

For the 357 Hz tuning fork, the vibrational frequency is;

f = 357 ± 8

f = (357 + 8) or (357 - 8)

f = 365 Hz or 349 Hz

In both cases, 349 Hz is common;

Thus, the vibrational frequency of the string = 349 Hz

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A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
Irina-Kira [14]

Answer:

part (a) \alpha\ =\ -2.5\ rad/s^2

part (b) N = 79.61 rev

part (c) \tau\ =\ 23.54\ Nm

Explanation:

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  • total time taken = t = 20.0 sec

part (a)

Let \alpha be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2

part (b)

Let \theta be the total angular displacement of the wheel from initial position till the rest.

\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad

We know,  1 revolution = 2\pi rad

Let N be the number of revolution covered by the wheel.

\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

  • Mass of the pole = m = 4 kg
  • Length of the pole = L = 2.5 m
  • Angle of the pole with the horizontal axis = \theta\ =\ 60^o

Now the center of mass of the pole = d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m

Weight component of the pole perpendicular to the center of mass = F\ =\ mgcos\theta

\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm

3 0
4 years ago
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