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In an experiment on standing waves, a string90 cmlong is attached to the prong of an electrically driven tuning fork that oscillates perpendicular to the length of the string at a frequency of 60 Hz. The mass of the string is 0.044 kg . What tension must the string be under (weights are attached to the other end) if it is to oscillate in four loops?

Short Answer

Expert verified

The tension in the string if it is to oscillate in four loops is 36 N .

Step by step solution

01

Given data

Length of string is,L=90cmor0.90m

The frequency of the tuning fork is,f=60Hz

Mass of the string is,m=0.044kg.

02

Understanding the concept of frequency of oscillation

We know the formula for the frequency of the standing wave to oscillate in n loops in terms of the velocity of the standing wave (v). We also know the formula for combining these two formulae. Rearranging them, we can get an expression for the tension in the string. Inserting the values in it, we can find its value.

Formula:

The frequency of standing wave for n loops of oscillation,f=nv2L.......1

The velocity of the body, v=Tμ.......2

03

Calculation of the tension of the string

Frequency of standing wave to oscillate in n loops

f=nv2L

Using equation (2), the velocity of the string can be given as:

v=TmL.........3(μ=mL)

Substituting the value of equation (3) in equation (1), we get the frequency of the oscillation as:

f=nTmL2Lf=nTmL2f2=n2TmL4T=4f2mLn2=46020.0440.9042=35.6436N

Therefore, the tension in the string if it is to oscillate in four loops is 36N.

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Most popular questions from this chapter

(a) If a standing wave on a string is given by

y(t)=(3mm)sin(5x)cos(4t)

is there a node or an antinode of the oscillations of the string atx = 0? (b) If the standing wave is given by

y(t)=(3mm)sin(5x+p/2)cos(4t)

is there a node or an antinode at x = 0?

The tension in a wire clamped at both ends is doubled without appreciably changing the wire’s length between the clamps. What is the ratio of the new to the old wave speed for transverse waves traveling along this wire?

A wave on a string is described by y(x,t)=15.0sin(πx/8-4πt), where xand yare in centimeters and tis in seconds. (a) What is the transverse speed for a point on the string at x = 6.00 cm when t = 0.250 s? (b) What is the maximum transverse speed of any point on the string? (c) What is the magnitude of the transverse acceleration for a point on the string at x = 6.00 cm when t = 0.250 s? (d) What is the magnitude of the maximum transverse acceleration for any point on the string?

In Figure 16-36 (a), string 1 has a linear density of 3.00 g/m, and string 2 has a linear density of 5.00 g/m. They are under tension due to the hanging block of mass M = 500 g. (a)Calculate the wave speed on string 1 and (b) Calculate the wave speed on string 2. (Hint:When a string loops halfway around a pulley, it pulls on the pulley with a net force that is twice the tension in the string.) Next the block is divided into two blocks (with M1+M2=M) and the apparatus is rearranged as shown in Figure (b). (c) Find M1and (d) Find M2such that the wave speeds in the two strings are equal.

A generator at one end of a very long string creates a wave given by

y1=(6.0cm)cosπ2[2.00m-1x+8.00s-1t]

and a generator at the other end creates the wave

y2=(6.0cm)cosπ2[(2.00m-1)x-(8.00s-1)t]

Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave.

For x0, what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of x? Forx0, what is the location of the antinode having the (g) smallest, (h) second smallest, and (i) third smallest value ofx?

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