Answer:
Explanation:
(A)
The string has set of normal modes and the string is oscillating in one of its modes.
The resonant frequencies of a physical object depend on its material, structure and boundary conditions.
The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.
Given below are the incorrect options about the wave in the string.
• The wave is travelling in the +x direction
• The wave is travelling in the -x direction
• The wave will satisfy the given boundary conditions for any arbitrary wavelength  
• The wave does not satisfy the boundary conditions 
Here, the string of length L held fixed at both ends, located at x=0 and x=L
The key constraint with normal modes is that there are two spatial boundary conditions,
 and 
.The spring is fixed at its two ends.
The correct options about the wave in the string is
• The wavelength  can have only certain specific values if the boundary conditions are to be satisfied.
  can have only certain specific values if the boundary conditions are to be satisfied.
(B)
The key factors producing the normal mode is that there are two spatial boundary conditions,  and
 and  , that are satisfied only for particular value of
, that are satisfied only for particular value of  .
  .
Given below are the incorrect options about the wave in the string.
•   must be chosen so that the wave fits exactly o the string.
 must be chosen so that the wave fits exactly o the string.
• Any one of   or
 or  or
  or  can be chosen to make the solution a normal mode.
  can be chosen to make the solution a normal mode.
Hence, the correct option is that the system can resonate at only certain resonance frequencies  and the wavelength
 and the wavelength  must be such that
  must be such that 
(C)
Expression for the wavelength of the various normal modes for a string is,
 (1)
 (1)
When  , this is the longest wavelength mode.
 , this is the longest wavelength mode.
Substitute 1 for n in equation (1).

When  , this is the second longest wavelength mode.
 , this is the second longest wavelength mode.
Substitute 2 for n in equation (1).

When  , this is the third longest wavelength mode.
, this is the third longest wavelength mode.
Substitute 3 for n in equation (1).

Therefore, the three longest wavelengths are  ,
, and
 and  .
.
(D)
Expression for the frequency of the various normal modes for a string is,

For the case of frequency of the  normal mode the above equation becomes.
 normal mode the above equation becomes.
 
Here,  is the frequency of the
 is the frequency of the  normal mode, v is wave speed, and
 normal mode, v is wave speed, and  is the wavelength of
 is the wavelength of  normal mode.
 normal mode.
Therefore, the frequency of  normal mode is
 normal mode is   
.