Hello can you do these problems in the attachments the answers are in the end of the questions i just want to know how i get the answer. Please use the Equation Sheet to answer the questions. I also...

1)
The tension 1T can be resolved along vertical and horizontal axis as shown in the figure .
1 2 1T sin30 T and T cos30 5x9.8  
We have
1
5x9.8
T N
cos30

and
2
5x9.8xsin30 1
T 49x tan30 49x 28.29 N
cos30 3
   
The speed of a wave “v” in the string T2 is given by
2
2
1
3
T
v ( whereT tension in thestring and linear densityof thestring)
28.29
59.5ms
8x10


   

 
2)
The general form of a wave equation is described below .
2 f (f frequencyof thewave) rad/ s  
2
k ( wavelength) rad/ m or / m

  


12v (velocityof the wave) f x ms (1)
k 2 k
      


We have the given wave equation as
y 0.2m sin (0.18rad/ m) x (90rad/ s) t
4
   
 

Now comparing this equation with the general equation , we have
2)
Amplitude of the wave is 0.2m
3)
wave length ,
2 2
34.9m ( k 0.18, as given in theequation)
k 0.18
 
    
4)
wave velocity 1
90
v 500ms ( from eq(1) )
k 0.18
   and travels towards LEFT ( + is in the eq.)
5)
How wave passes by a point on the string per second , this is nothing but frequency “f” (ie no of oscillations or
passes per second )
90
f 14.32 14.3 waves / sec
2 2

   
 

6)
we obtain displacement at t=0 and x=0 , by substituting t=0 and x=0 in the given wave eq.
y( 0.2m sin (0.18rad/ m)0 (90rad/ s)0 0.2m sin[ ] 0.141m
4
t 0 and x 0)
4
        

  


7)
Given mass per uint length (linear density ) µ = 6g/m =6x10-3 kg/m
The energy per sec [or power] is given by
2 2
av
1
P A v
2
 
Substituting the values from the above equations , we get
3 2 2
av
1
P x60x10 x90 x(0.2) x500 486 W
2
 
8)
The radio waves are electromagnetic waves which travels with the speed of light, and is given by
8
6
c where wavelength
frequency
c speed of ligth
Hence,
c 3x10
3.09m
97x10
   
 

   


9)
In Doppler effect , we follow the sign convention as shown below.
when the source and the observer are moving towards each other , then the relationship between emitted
frequency by source and the received frequency by the Observer is given by
sound obseve
o
sound source
V V
f f
V V
 
  
 

Where fo = received frequency by the Observer
f= emitted frequency by resource
V= velocity of (sound/source/Observer as the subscript)
Here ,
the velocity of sound is given as 1533 m/s
and let’s say source is travelling with a velocity of “v” , then the velocity of Observer is “2v” , as per the
given condition in the problem .
sound
source
obseve
V 1533m / s
V v
V 2v
 



Substituting the values in the above equation, we get .
1
1533 2v
302 299
1533 v
302
x(1533 v) 1533 2v
299
302
x1533 1533
299
v
302
2
299
v 5.110000 5.11ms (speed of source / Humphery)
 
  
 
 
   
 
 
 
  
 
 
 
 

10)
The harmonic for an open pipe is given by
n
nv
f , n 1,2,3...
2L
 
Where fn = frequency of the nth harmonic
L= length of the pipe
V= velocity of sound in the pipe
2
2v
f
2L
  and 3...