TG EAMCET 4 May 2025 Shift 1 Paper

The tube is 50 cm long and is closed at one end.
Finding the Wavelength:
For the fifth harmonic in a closed pipe, the formula is:
‌

5λ

4

=L
This means five-fourths of the wavelength fits into the total length (L) of the tube.
Let's solve for wavelength (λ) :
λ=‌

4L

5

=‌

4×50

5

=40cm
Calculating the Phase Difference:
The phase difference (∆φ) between two points separated by a distance x is:
∆φ=‌

2π

λ

×x
Here, x=42cm (the distance from the open end to the other particle) and λ=40cm :
∆φ=‌

2π

40

×42
This can be calculated as:
∆φ=‌

2π×42

40

=‌

84π

40

=‌

21π

10

If you convert this to degrees:
∆φ=‌

21π

10

×‌

180∘

π

=378∘

But phases repeat every 360∘. So, we subtract 360∘ to get the simpler value:
378∘−360∘=18∘
So, the phase difference is 18∘.