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#8 | Solved Problem on Speed of Transverse Wave
(Physics) > Waves

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A uniform rope of length \(L\) and mass \(m_1\) hangs vertically from a rigid support. A block of mass \(m_2\) is attached to the free end of the ropes. A transverse pulse of wavelength \(\lambda_1\) is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is \(\lambda_2\). The ratio \(\dfrac{\lambda_2}{\lambda_1}\) is:
1. \(\sqrt{\dfrac{m_1+m_2}{m_1}}\)
2. \(\sqrt{\dfrac{m_2}{m_1}}\)
3. \(\sqrt{\dfrac{m_1+m_2}{m_2}}\)
4. \(\sqrt{\dfrac{m_1}{m_2}}\)

From NCERT
NEET - 2016
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If n1, n2 and n3 are, are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by

1. 1/n=1/n1+1/n2+1/n3
 

2. 1/√n=1/√n1+1/√n2+1/√n3

3. √n=√n1+√n2+√n3

4. n=n1+n2+n3

 70%
From NCERT
NEET - 2014
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When a string is divided into three segments of lengths l1,l2andl3, the fundamental frequencies of these three segments are v1,v2andv3 respectively. The original fundamental frequency (v) of the string is 

1. v=v1+v2+v3

2. v=v1+v2+v3

3. 1v=1v1+1v2+1v3

4. 1v=1v1+1v2+1v3

 79%
From NCERT
NEET - 2012
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A wave in a string has an amplitude of 2 cm. The wave travels in the +ve direction of x-axis with a speed of 128ms-1 and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is :

1. y=0.02msin7.85x+1005t

2. y=0.02msin15.7x-2010t

3. y=0.02msin15.7x+2010t

4. y=0.02msin7.85x-1005t

 68%
From NCERT
NEET - 2009
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The equation of a wave traveling in a string can be written as y=3cosπ(100tx). Its wavelength is :

(1) 100 cm

(2) 2 cm

(3) 5 cm

(4) None of the above

 83%
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