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Engineering
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If n1, n2 and n3 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:

• Option 1)

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

• Option 2)

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

• Option 3)

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

• Option 4)

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

As we learnt in  Fundamental frequency with end correction -     (one end open)     (Both end open) e = end correction -   Option 1) Correct Option 2) Incorrect Option 3) Incorrect Option 4) Incorrect
Engineering
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Two sources of sound placed close to each other are emitting progressive waves given by y1 = 4 sin 600 t and y2 = 5 sin 608 t. An observer located near these two sources of sound will hear:

• Option 1)

4 beats per second with intensity ratio 25: 16 between waxing and waning.

• Option 2)

8 beats per second with intensity ratio 25 : 16 between waxing and waning.

• Option 3)

8 beats per second with intensity ratio 81 : 1 between waxing and waning.

• Option 4)

4 beats per second with intensity ratio 81 : 1 between waxing and waning.

As we learnt in Beat Frequency - module of   - wherein Where are frequency of two wave differ slightly  in value of frequency.     Ratio of maximum and minimum intensity - -                                 ....................(i)                             ....................(ii) Correct option is 4.     Option 1) 4 beats per second with intensity ratio 25: 16 between waxing and...
Engineering
60 Views   |

When a string is divided into three segments of length , and  the fundamental frequencies of these three segments are v1, v2 and v3 respectively. The original fundamental frequency (v) of the string is

• Option 1)

• Option 2)

v = v1+ v+ v3

• Option 3)

• Option 4)

$\frac{1}{\sqrt{\text{v}}}=\frac{1}{\sqrt{\text{v}_{1}}}+\frac{1}{\sqrt{\text{v}_{2}}}+\frac{1}{\sqrt{\text{v}_{3}}}$

Engineering
539 Views   |

The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are:

• Option 1)

Kgms-1

• Option 2)

Kgms-2

• Option 3)

Kgs-1

• Option 4)

Kgs

As we discussed in Option 1) Kgms-1 This option is incorrect. Option 2) Kgms-2 This option is incorrect. Option 3) Kgs-1 This option is correct. Option 4) Kgs This option is incorrect.
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