मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

The Fundamental Frequency of a String is Proportional to

Advertisements
Advertisements

प्रश्न

The fundamental frequency of a string is proportional to

पर्याय

  • inverse of its length

  • the diameter

  •  the tension

  • the density.

MCQ
Advertisements

उत्तर

inverse of its length

The relation between wave speed and the length of the string is given by

\[v = \frac{1}{2l}\sqrt{\frac{F}{\mathrm{\mu}}}\] 
where
           l is the length of the string
           is the tension
           μ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.

\[v \propto \frac{1}{l}\]

shaalaa.com
Speed of Wave Motion
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 15: Wave Motion and Waves on a String - MCQ [पृष्ठ ३२२]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 15 Wave Motion and Waves on a String
MCQ | Q 18 | पृष्ठ ३२२

संबंधित प्रश्‍न

Both the strings, shown in figure, are made of same material and have same cross section. The pulleys are light. The wave speed of a transverse wave in the string AB is
\[\nu_1\]  and in CD it is \[\nu_2\]. Then \[\nu_1 / \nu_2\]


Two periodic waves of amplitudes A1 and A2 pass thorough a region. If A1 > A2, the difference in the maximum and minimum resultant amplitude possible is


A wave pulse passing on a string with a speed of 40 cm s−1 in the negative x-direction has its maximum at x = 0 at t = 0. Where will this maximum be located at t = 5 s?


following Figure shows a wave pulse at t = 0. The pulse moves to the right with a speed of 10 cm s−1. Sketch the shape of the string at t = 1 s, 2 s and 3 s.


A wave is described by the equation \[y = \left( 1 \cdot 0  mm \right)  \sin  \pi\left( \frac{x}{2 \cdot 0  cm} - \frac{t}{0 \cdot 01  s} \right) .\] 
(a) Find the time period and the wavelength? (b) Write the equation for the velocity of the particles. Find the speed of the particle at x = 1⋅0 cm at time t = 0⋅01 s. (c) What are the speeds of the particles at x = 3⋅0 cm, 5⋅0 cm and 7⋅0 cm at t = 0⋅01 s?
(d) What are the speeds of the particles at x = 1⋅0 cm at t = 0⋅011, 0⋅012, and 0⋅013 s?


A string of linear mass density 0⋅5 g cm−1 and a total length 30 cm is tied to a fixed wall at one end and to a frictionless ring at the other end (See figure). The ring can move on a vertical rod. A wave pulse is produced on the string which moves towards the ring at a speed of 20 cm s−1. The pulse is symmetric about its maximum which is located at a distance of 20 cm from the end joined to the ring. (a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to region its shape. (b) The shape of the string changes periodically with time. Find this time period. (c) What is the tension in the string?


The speed of sound in a medium depends on


A one-metre long stretched string having a mass of 40 g is attached to a tuning fork. The fork vibrates at 128 Hz in a direction perpendicular to the string. What should be the tension in the string if it is to vibrate in four loops?


In Quincke's experiment the sound detected is changed from a maximum to a minimum when the sliding tube is moved through a distance of 2.50 cm. Find the frequency of sound if the speed of sound in air is 340 m s−1.


Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. Speed of sound in air is 340 ms−1.


A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. Speed of sound in air = 340 m s−1.


In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20.0 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form?


A copper rod of length 1.0 m is clamped at its middle point. Find the frequencies between 20 Hz and 20,000 Hz at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is 3.8 km s−1.


A train approaching a platform at a speed of 54 km h−1 sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. the train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platform? The speed of sound in air = 332 m s−1.


A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note. When the two are closed by and the train approaches the person on the ground, he hears 4.0 beats per second. The speed of sound in air = 340 m s−1. (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train?


Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km h−1 and the other at 54 km h−1 relative to the water. The first submarine sends a sound signal (sound waves in water are also called sonar) at a frequency of 2000 Hz. (a) At what frequency is this signal received from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of of the sound wave in water to be 1500 m s−1.


A source emitting a sound of frequency v is placed at a large distance from an observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is v.


A metallic wire of 1 m length has a mass of 10 × 10−3 kg. If the tension of 100 N is applied to a wire, what is the speed of the transverse wave?


A spring breaks under tension of 10 kg wt.If the string is used to revolve a body of mass 1.2 kg in a horizontal circle. of radius 50 cm, what is the maximum speed with which a body can be revolved?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×