A standing wave is set up in a string of variable length and...

A standing wave is set up in a string of variable length and tension by a vibrator of variable frequency. Both ends of the string are fixed. When the vibrator has a frequency $\int_{A}$, in a string of length $L_{A}$ and under tension $T_{A}, n_{A}$ antinodes are set up in the string. (a) Write an expression for the frequency $f_{A}$ of a standing wave in terms of the number $n_{A}$, length $L_{d}$, tension $T_{A}$, and linear density $\mu_{i} \cdot$ (b) If the length of the string is doubled to $L_{l l}=2 L_{i}$, what frequency $f_{B}$ (written as a multiple of $f_{A}$ ) will result in the same number of antinodes? Assume the tension and linear density are unchanged. Hint: Make a ratio of expressions for $f_{B}$ and $f_{A}$. (c) If the frequency and length are held constant, what tension $T_{B}$ will produce $n_{A}+1$ antinodes? (d) If the frequency is tripled and the length of the string is halved. by what factor should the tension be changed so that twice as many antinodes are produced?

Transcript

00:01 As shown in the figure, this is the first mode of vibration.

00:08 A string is tied between two ends, right? so when string vibrates, the nodes are generated at the two ends of the string where string is fixed because there is no vibration.

00:22 An antinode is formed at the center of the string where the vibration is maximum.

00:28 This is the first mode of vibration of the string where we have number of n -node n -a is equal to 1.

00:39 Now from the standard formula, the frequency of vibration in the first mode is given by 1 upon 2la, which is the length of the string multiplied by under root t .a.

01:02 Tension in the string and mu i which is the mass per unit length of the string now in second case where the string vibrates in second mode we have two anti -nodes are formed right two anti -notes are formed here at the center and we have three notes so here number of antinodes is equal to two and the frequency of vibration is given as 2 times 1 upon 2la under root t by mu i these are the standard formula of standing waves in second mode of vibration where we have 3 antinodes n -a equal to 3 as per the standard formula the frequency of vibration is given as three times 1 upon 2la under root t upon mu i right now if we come if we see equation 1 2 and 3 if we try to find some similarity equation 1 if we say that equation 1 is here multiplied by 1 then equation 2 f2 is the same value multiplied by 2 and equation 3 has the same value these values are same and here it is multiplied by 3 so it means in 1 here it is multiplied by 1 then 2 by 3 now if we see what is 1 1 is the number of antinot here what is 2 2 is the number of antinot 2 what is 3 here 3 means 3 is the number of antinodes.

03:13 So in this way, we can journalize these equation 1, 2 and 3 for any number of nodes, we can mark fn is equal to n a upon 2la under root t .a by mui.

03:43 These are this is the standard equation of frequency of standing waves where n a is the number of antinodes la is the length of the string t a is the tension and mua is the mass per unit length right this is part a of the same problem now we will move to part b in part b they are saying the number of antinodes remains the same but length of the string is doubled so how the how the frequency should change right so we will write two equation first equation is the number of antinode n a is equal to the frequency f a same means from this same is from this equation we can write n a as f a into two la under root mu i upon t a same is from this equation we can find out the value of n a n a is equal to f a multiplied by 2 n a multiplied by under root mu i by t a now this is the first case of antinodes now in second case let number of antinode is n b n b is equal to f b into 2 l b under root mu i upon t b say this is mu i a this is mu i b mass per unit length now in the given problem they are saying that n a equal to n b means number of antinode remains the same it means f a multiplied by 2 l a under root m a under root mu i sorry under root mu i sorry under root mu i a under root mu i a upon t a is equal to n b n b is nothing but f b to l b under root mu i b upon t b now in the given given problem they are saying that the tension and the linear density remains same it means what t a equal to means t b they will cancel out and the linear density remains same it means mu i a will cancel out with mu i b right so and two and two are also cancelled out right so now they are saying the length of the thing is doubled so we can write here again f a multiplied by la f b now f b now l lb is 2 times la, say because length is doubled.

07:37 Right.

07:37 Here again, la and la cancelled out.

07:41 La cancelled out with la.

07:44 So we can write here finally fb is equal to fa by 2.

07:53 It means the frequency will be half as compared to the initial frequency.

08:03 Then this is part b now we'll come to part c in part c they are saying the number of antinode is increased by 1 if the frequency and length are held constant how the tension will change if the number of antinodes are increased by 1 so here where f and l are constant.

08:50 So we can write n -a is equal to f -a -2 -l -a under root mu -i upon t -a.

09:09 And n -b, after making the changes, let number of anti -nodes is n -b, n -b is equal to f -b to l -b under root mu i a mu i b upon t b right now if we divide these two equations if let this is be equation one let this is be equation two if we divide these two equations then what we get is n a upon n b n a upon n b is equal to f a upon f b multiplied by 2 l a upon 2 l b multiplied by under root mu i a under root mu i a under root mu i a upon mu i b mass per mass per unit length in two different condition multiplied by t b tension upon t a right now here they are saying that the frequency remains same f a equal to fb they are also saying that the length remains the same la equal to lb now if length remains the same it means mass per unit length also remains the same because it is the same string right so it means mu i a should be equal to mu i b right so if we see here mu i a and mu i b will be cancelled out similarly l a l a l la will be cancelled out and similarly we have faa f and fb will also cancel out.

11:33 What they are saying is nb should be equal to n a plus 1.

11:42 Say because they are saying what tension will produce n a plus 1 antinodes...

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