Two sound waves expressed in CGS units given by y 1-0-3 sin2 - vt - x and y 2-0-4 sin2 - vt - x - interfere- The resultant amplitudeat a place where phase difference is -2 will be A- 0-1 cmB- 0-5 cmC- 1-10-7 cmD- 0-7 cm

The correct option is B 0.5 cmResultant amplitude =√(a_1^2+a_2^2+a_1a_2 cos ϕ) =√(0.3^2+0.4^2+2× 0.3× 0.4× cosπ/2)=0.5 cm ...

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Standard XII

Physics

YDSE Problems II

Two sound wav...

Question

Two sound waves (expressed in CGS units) given by $y_{1} = 0.3 s i n \frac{2 π}{λ} (v t - x)$ and $y_{2} = 0.4 s i n \frac{2 π}{λ} (v t - x + θ)$ interfere. The resultant amplitude at a place where phase difference is $\frac{π}{2}$ will be

0.7 cm

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0.1 cm

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0.5 cm

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Solution

The correct option is C 0.5 cm
Resultant amplitude $= \sqrt{a_{1}^{2} + a_{2}^{2} + a_{1} a_{2} c o s ϕ}$

$= \sqrt{{0.3}^{2} + {0.4}^{2} + 2 \times 0.3 \times 0.4 \times c o s \frac{π}{2}} = 0.5$ cm

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Two sound waves (expressed in CGS units) given by y1=0.3 sin2πλ(vt−x) and y2=0.4 sin2πλ(vt−x+θ) interfere. The resultant amplitude at a place where phase difference is π2 will be

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Two sound waves (expressed in CGS units) given by $y_{1} = 0.3 s i n \frac{2 π}{λ} (v t - x)$ and $y_{2} = 0.4 s i n \frac{2 π}{λ} (v t - x + θ)$ interfere. The resultant amplitude at a place where phase difference is $\frac{π}{2}$ will be

The correct option is C 0.5 cm
Resultant amplitude $= \sqrt{a_{1}^{2} + a_{2}^{2} + a_{1} a_{2} c o s ϕ}$

$= \sqrt{{0.3}^{2} + {0.4}^{2} + 2 \times 0.3 \times 0.4 \times c o s \frac{π}{2}} = 0.5$ cm