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. \(\sqrt{\dfrac{m_1}{m_2}}\)

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

When a string is divided into three segments of lengths \(l_1,~l_2\text{ and }l_3,\) the fundamental frequencies of these three segments are \(\nu_1,~\nu_2\text{ and }\nu_3\) respectively. The original fundamental frequency \((\nu)\) of the string is:
1. \(\sqrt{\nu}=\sqrt{\nu_1}+\sqrt{\nu_2}+\sqrt{\nu_3}\)
2. \(\nu=\nu_1+\nu_2+\nu_3\)
3. \(\dfrac{1}{\nu}=\dfrac{1}{\nu_1}+\dfrac{1}{\nu_2}+\dfrac{1}{\nu_3}\)
4. \(\dfrac{1}{\sqrt{\nu}}=\dfrac{1}{\sqrt{\nu_1}}+\dfrac{1}{\sqrt{\nu_2}}+\dfrac{1}{\sqrt{\nu_3}}\)

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 $128<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-1}$ and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is :

1. $y=\left(0.02\right)\mathrm{m}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\left(7.85\mathrm{x}<mspace\; width="1em"></mspace>+1005\mathit{\hspace{0.17em}}t<mspace\; width="1em"></mspace>\right)$

2. $y=\left(0.02\right)\mathrm{m}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\left(15.7\mathrm{x}<mspace\; width="1em"></mspace>-2010\mathit{\hspace{0.17em}}t<mspace\; width="1em"></mspace>\right)$

3. $y=\left(0.02\right)\mathrm{m}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\left(15.7\mathrm{x}<mspace\; width="1em"></mspace>+2010\mathit{\hspace{0.17em}}t<mspace\; width="1em"></mspace>\right)$

4. $y=\left(0.02\right)\mathrm{m}<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\left(7.85\mathrm{x}<mspace\; width="1em"></mspace>-1005\mathit{\hspace{0.17em}}t<mspace\; width="1em"></mspace>\right)$

The equation of a wave traveling in a string can be written as $y=3\cos \pi (100t-x)$. Its wavelength is :

(1) 100 cm

(2) 2 cm

(3) 5 cm

(4) None of the above

A string fixed at both ends is vibrating in two segments. The wavelength of the corresponding wave is :

(1) $\frac{l}{4}$

(2) $\frac{l}{2}$

(3) l

(4) 2l

A 1 cm long string vibrates with the fundamental frequency of 256 Hz. If the length is reduced to $\frac{1}{4}cm$  keeping the tension unaltered, the new fundamental frequency will be :

(1) 64

(2) 256

(3) 512

(4) 1024

A string is producing transverse vibration whose equation is $y=0.021\sin (x+30t)$, Where x and y are in meters and t is in seconds. If the linear density of the string is 1.3×10^–4 kg/m, then the tension in the string in N will be :

(1) 10

(2) 0.5

(3) 1

(4) 0.117

A string on a musical instrument is 50 cm long and its fundamental frequency is 270 Hz. If the desired frequency of 1000 Hz is to be produced, the required length of the string is :

(1) 13.5 cm

(2) 2.7 cm

(3) 5.4 cm

(4) 10.3 cm

The tension in a piano wire is 10N. What should be the tension in the wire to produce a note of double the frequency :

(1) 5 N

(2) 20 N

(3) 40 N

(4) 80 N