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A 2.0-m-long string vibrates at its second-harmonic frequency with a maximum amplitude of 2.0 cm. One end of the string is at x = 0 cm. Find the oscillation amplitude at x = 10, 20, 30, 40, and 50 cm.

Short Answer

Expert verified

The amplitude of the wave at x =10 cm, 20cm, 30cm, 40cm and 50cm is 6.2mm, 11.8mm, 16.2 mm, 19.0 mm, 19.0 mm and 20mm respectively

Step by step solution

01

Write the given information 

Length of string L= 2m
Second harmonic frequency, n= 2
The amplitude of the wave= 2 cm = 0.02 m

02

 Step 2: To determine the wavelength of the string wave 

Let the wavelength is denoted by 位

The wavelength of the nth harmonic frequency is given by
n=2Ln


Here n=2,

2=2L2=L2=2m
Thus, the wavelength of the wave is 2m

03

To determine the amplitude wave at different positions 

The general expression of the amplitude of a wave is given by
A(x)=2asin2x

Substitute the known values

A(x)=0.02sin2x2A(x)=0.02sinx......(1)

Substitute x= 10 cm=0.1m in equation (1)
A(0.1)=(0.02)sin((0.1))=6.2mm

Put x= 20cm = 0.2m in equation (1)
A(0.2)=(0.02)sin((0.2))=11.8mm

Put x= 30cm = 0.3m in equation (1)
A(0.3)=(0.02)sin((0.3))=16.2mm

Put x= 40cm = 0.4m in equation (1)
A(0.4)=(0.02)sin((0.4))=19.0mm

Put x= 50cm = 0.5m in equation (1)
A(0.5)=(0.02)sin((0.5))=20.0mm

Thus, the amplitude of the wave at x =10 cm, 20cm, 30cm, 40cm and 50cm is 6.2mm, 11.8mm, 16.2 mm, 19.0 mm, 19.0 mm and 20mm respectively

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