PHYSICS COLLEGE

The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.

Answers

Answer 1

Answer:

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_(beat) = 0.99s$

Generally the frequency of the beat is

$f_(beat) = (1)/(t_(beat))$

Substituting values

$f_(beat) = (1)/(0.99)$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_(beat)$ for $f_2$ having a higher tension

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((231.01)^2)/((230)^2)$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_(beat)$

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((228.99)^2)/((230)^2)$

$T_2 = 0.99$ % lower than $T_1$

Related Questions

A virtual image produced by a lens is always
A block of ice with mass 2.00 kg slides 0.775 m down an inclined plane that slopes downward at an angle of 31.8 ∘ below the horizontal. A) If the block of ice starts from rest, what is its final speed? You can ignore friction.
A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?
Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is four times greater than that of wire B. What is the ratio of the radius of A to that of B
The heating element of a coffeemaker operates at 120 V and carries a current of 4.50 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.525 kg of water from room temperature (23.0°C) to the boiling point.

The tension of a string is found to be 4050 N and the mass of 1 m of the string is 0.5 kg. What could be the velocity of a wave traveling in that string?

Answers

Answer:

90 m/s

Explanation:

The formula for velocity of wave in a string is given as,

v' = √(T/m') ................ Equation 1

Where v' = velocity of the string, T = Tension on the string, m' = mass per unit length of the string.

Given: T = 4050 N, and

m' = m/l where m = mass of the string, l = length of the spring.

m = 0.5 kg, l = 1 m

m' = 0.5/1 = 0.5 kg/m

Substitute into equation 1

v' = √(4050/0.5)

v' = √(8100)

v' = 90 m/s.

Hence the velocity of the wave in the string = 90 m/s

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail. When small prey, or even potential mates, swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.00 μN/C. How much charge, modeled as a point charge, in the fish would be needed to produce such a change in the electric field at a distance of 63.5 cm ?

Answers

Answer:

$1.34\cdot 10^(-16) C$

Explanation:

The strength of the electric field produced by a charge Q is given by

$E=k(Q)/(r^2)$

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

$E=3.00 \mu N/C = 3.00\cdot 10^(-6)N/C$

and the fish can detect the electric field at a distance of

$r=63.5 cm = 0.635 m$

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

$Q=(Er^2)/(k)=((3.00\cdot 10^(-6) N/C)(0.635 m)^2)/(9\cdot 10^9 Nm^2 C^(-2))=1.34\cdot 10^(-16) C$

Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and friction acts on the body in the opposite direction with a force of 1 N, the net force would be 9 N in the direction of the spring balance (10 N – 1 N = 9 N).What is the net force acting on the object when the spring balance pulls the rope with a force of 25 N and friction acts on the body with a force of 20N?

Answers

Answer:

Explanation:

(25 N - 20 N = 5 N)

A beam of light travels from a medium with an index of refraction of 1.27 to a medium with an index of refraction of 1.46. If the incoming beam makes an angle of 14.0° with the normal, at what angle from the normal will it refract?

Answers

Answer:

12.15°

Explanation:

Using Snell's law as:

$n_i* {sin\theta_i}={n_r}*{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 14.0° )

${\theta_r}$ is the angle of refraction ( ? )

${n_r}$ is the refractive index of the refraction medium (n=1.46)

${n_i}$ is the refractive index of the incidence medium (n=1.27)

Hence,

$1.27* {sin14.0^0}={1.46}*{sin\theta_r}$

Angle of refraction = $sin^(-1)0.2104$ = 12.15°

Answer:

Explanation:

In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the slower runner when the faster runner finishes the race?

Answers

Answer:

The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

Explanation:

Given;

the speed of the slower runner, u₁ = 11.8 km/hr

the speed of the fastest runner, u₂ = 15 km/hr

distance, d = 8 km

The time when the fastest runner finishes the race is given by;

$Time = (Distance )/(speed)\n\nTime = (8)/(15) \n\nTime = 0.533 \ hr$

The distance covered by the slower runner at this time is given by;

d₁ = u₁ x 0.533 hr

d₁ = 11.8 km/hr x 0.533 hr

d₁ = 6.29 km

Additional distance (x) the slower runner need to finish is given by;

6.29 km + x = 8km

x = 8 k m - 6.29 km

x = 1.71 km

Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the projectile a freely falling body?Yes or No(b) What is its acceleration in the vertical direction? (Let up be the positive direction.)____? m/s2(c) What is its acceleration in the horizontal direction?

Answers

Answer:

A) No

B)-9,81 m/s^2

C)0 m/s^2

Explanation:

A free fallin object has only velocity on the vertical axis so any object that is moving in the Y and X axis has projetile motion not free falling, and when dealing with projectile motion the object is experiencing acceleration towards the ground of -9,81m/s^2 and in the Y axis, in the X axis there´s is only acceleration if the air is providing resistance, since it states that it isnot, then the accleration is 0.

Other Questions

It's a little hard can u help

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?

A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total distance 25.0 mi at 52.00 mph. (a) What is the total time in hours of the trip? Keep two decimal places. 4.05 Correct (100,0%) (b) What is the car's average speed in mph for the entire trip? Keep two decimal places. 12 35 Correct (100,0%) Submit The car travels the same distance again, but this time, in the first half of the time its speed is 7.00 mph and in the second half of the time its speed is 52.00 mph. c) What is the total time in hours of the trip? Keep two decimal places.

A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of 60 cm. Find the work done in stretching the spring from 60 cm to 65 cm. First, setup an integral and find a a, b b, and f ( x ) f(x) which would compute the amount of work done.