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The first overtone of a stretched wire of given length is 320 Hz. The first harmonic is :

320 Hz

160 Hz

480 Hz

640 Hz

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Explanation

Frequency of first overtone or second harmonic (n₂) = 320 Hz. So, frequency of first harmonic $n_{1} = \frac{n_{2}}{2} = \frac{320}{2} = 160 H z$

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The sound carried by the air from a sitar to a listener is a wave of the following type :

Longitudinal stationery

Transverse progressive

Transverse stationery

Longitudinal progressive

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Explanation

Observer receives sound waves (music) which are longitudinal progressive waves.

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Three similar wires of frequency n₁, n₂ and n₃ are joined to make one wire. Its frequency will be :

$n = n_{1} + n_{2} + n_{3}$

$\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}$

$\frac{1}{\sqrt{n}} = \frac{1}{\sqrt{n_{1}}} + \frac{1}{\sqrt{n_{2}}} + \frac{1}{\sqrt{n_{3}}}$

$\frac{1}{n^{1}} = \frac{1}{n_{1}^{2}} + \frac{1}{n_{2}^{2}} + \frac{1}{n_{3}^{2}}$

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Explanation

$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$ ⇒ $n_{1} l_{1} = n_{2} l_{2} = n_{3} l_{3} = k$

$l_{1} + l_{2} + l_{3} = l$ ⇒ $\frac{k}{n_{1}} + \frac{k}{n_{2}} + \frac{k}{n_{3}} = \frac{k}{n}$

⇒ $\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + . .......$

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Two vibrating strings of the same material but lengths L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency n₁ and the other with frequency n₂. The ratio n₁/n₂ is given by :

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Explanation

Fundamental frequency $n = \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} ρ}}$

where m = Mass per unit length of wire

⇒ $n \propto = \frac{1}{l r} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{r_{2}}{r_{1}} \times \frac{l_{2}}{l_{1}} = \frac{r}{2 r} \times \frac{2 L}{L} = \frac{1}{1}$

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A string is rigidly tied at two ends and its equation of vibration is given by $y = (\cos 2 π t) (\sin 2 π x)$ Then minimum length of the string is :

1 m

$\frac{1}{2} m$

5 m

2π m

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Explanation

Given equation of stationary wave is

$y = \sin 2 π x \cos 2 π t$ ,

comparing it with standard equation $y = 2 A \sin \frac{2 π x}{λ} \cos ω t$

We have $\frac{2 π x}{λ} = 2 π x$ ⇒ $λ = 1 m$

Minimum length of string (first mode) , $L_{\min} = \frac{λ}{2} = \frac{1}{2} m$

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A string of length 2 m is fixed at both ends. If this string vibrates in its fourth normal mode with a frequency of 500 Hz then the waves would travel on its with a velocity of :

125 m/s

250 m/s

500 m/s

1000 m/s

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Explanation

For string $λ = \frac{2 l}{p}$

where p = No. of loops = Order of vibration

Hence for forth mode p = 4 ⇒ $λ = \frac{l}{2}$

Hence v = nλ $= 500 \times \frac{2}{2} = 500 H z$

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The fundamental frequency of a sonometre wire is n. If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be :

$\frac{n}{\sqrt{2}}$

$\frac{n}{2}$

$\frac{n}{2 \sqrt{2}}$

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Explanation

$n = \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} ρ}}$ ⇒ $n \propto \frac{\sqrt{T}}{r}$

⇒ $\frac{n_{2}}{n_{1}} = \frac{r_{1}}{r_{2}} \sqrt{\frac{T_{2}}{T_{1}}}$ $= \frac{1}{2} \times \sqrt{\frac{1}{2}} = \frac{1}{2 \sqrt{2}}$

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In an experiment with a sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. The tension of the string remaining constant the frequency of the second tuning fork is :

163.84 Hz

400 Hz

320 Hz

204.8 Hz

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Explanation

In case of sonometer frequency is given by

$n = \frac{p}{2 l} \sqrt{\frac{T}{m}}$ ⇒ $\frac{n_{2}}{n_{1}} = \frac{l_{1}}{l_{2}} \Rightarrow n_{2} = \frac{25}{16} \times 256 = 400 H z$

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The length of two open organ pipes are l and $(l + Δ l)$ respectively. Neglecting end correction, the frequency of beats between them will be approximate :

$\frac{v}{2 l}$

$\frac{v}{4 l}$

$\frac{v Δ l}{2 l^{2}}$

$\frac{v Δ l}{l}$

(Here v is the speed of sound)

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Explanation

$λ_{1} = 2 l, λ_{2} = 2 l + 2 Δ l$ ⇒ $n_{1} = \frac{v}{2 l}$ and $n_{2} = \frac{v}{2 l + 2 Δ l}$

⇒ No. of beats $= n_{1} - n_{2} = \frac{v}{2} (\frac{1}{l} - \frac{1}{l + Δ l}) = \frac{v Δ l}{2 l^{2}}$

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A tube closed at one end and containing air is excited. It produces the fundamental note of frequency of 512 Hz. If the same tube is open at both the ends the fundamental frequency that can be produced is :

1024 Hz

512 Hz

256 Hz

128 Hz

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Explanation

The fundamental frequency of open pipe is double that of the closed pipe.

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