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qaws [65]
3 years ago
12

Two point sources, vibrating in phase, produce an interferencepattern in a ripple tank. If the frequency is increased by 20%, th

enumber of nodal lines:
A) is increased by 20%
B) is increased by 40%
C) remains the same
D) is decreased by 20%
E) is decreased by 40%
Physics
1 answer:
Vera_Pavlovna [14]3 years ago
8 0

Answer:

option A

Explanation:

given,

frequency is increased by 20%

we know,

\dfrac{x_n}{L}d = (n-\dfrac{1}{2})\times \lambda...........(1)

where

x_n is the perpendicular distance between the point the interference pattern is obtained,

L is the distance between the center of the two point sources

and λ is  the wavelength of light.

If the frequency is increased by 20%, then the number of nodal lines is increased by 20%.

From equation (1),we observe that the frequency is directly proportional to the number of nodal lines.

Hence, the correct answer is option A

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Read 2 more answers
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
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