Home 1st Year Physics Notes FSC 1st Year Physics Chapter 8 Exercise & Numericals Notes

FSC 1st Year Physics Chapter 8 Exercise & Numericals Notes

747

In this post, I am sharing FSC 1st Year Physics Chapter 8 Solved Exercise & Numericals Notes PDF for the students of Intermediate Part 1. Waves is the name of this chapter. Students can download these 11th class physics chapter 8 Waves Solved Exercise & Numericals Notes PDF on their laptops or mobile. These Physics Notes for Class 11 are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. I have shared already complete Physics Notes for Class 11 of all chapters free.

FSC Part 1 Physics Chapter 8 Waves Solved Exercise & Numericals Notes PDF Download

Two speakers are arranged as shown in Fig. The distance between them is 3m and they emit a constant tone of 344Hz. A microphone P moved along a line parallel to and 4.00m from the line connecting the two speakers. It found that the tone of maximum loudness is heard and displayed on the CRO when the microphone on the Centre of the line and directly opposite each speaker. Calculate the speed of sound.

Data:

                Tone frequency = f = 344Hz
                distance between speaker = 3m
                distance between speaker and line of motion P = 4m
                speed of sound = v=?

Calculations;

For the tone of maximum loudness or constructive interference the path difference. At mid-point O path difference is zero, hence constructive interference takes place. For the next point of constructive interference, the path difference between two sound waves reaching points p1 and p2 should be equal to 1l, which can be calculated from the figure.
Path difference = S2P1 – S1P2

From the right angle triangle S2P1P2
S2P1  = √ (S2P1 ­)2– (S1P2)2
 
by putting value we have
  S2P1  = 5

Thus, the path difference = 5-4 =1m
as the path difference should be equal to 1l for constructive interference

Hence,

                     l = 1m
now using the relation,
v = fl
substituting the values we have
v = 344 x1

v = 344ms-1 (Ans)

For the remaining FSC 1st Year Physics Chapter 8 Solved Exercise & Numericals Notes PDF see the red pdf download button above. Please follow our Facebook page for educational updates.

Related Links

  1. 1st Year physics chapter 1 solved numericals notes pdf
  2. FSC Part 1 Year Physics Chapter 2 Solved Numericals Notes PDF
  3. 11th class Physics Chapter 3 Numericals Notes
  4. 11th Class Physics Chapter 4 Numericals notes pdf download
  5. 1st year physics chapter 5 Numericals Notes pdf download
  6. FSC 1st Year Physics Chapter 6 Notes pdf Download
  7. 1st year physics chapter 7 Notes pdf download
Load More In 1st Year Physics Notes

Leave a Reply

Your email address will not be published. Required fields are marked *