Numerical Problems Vibration of String Set-03

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In this article, we shall study to solve problems to calculate fundamental frequency of vibration of string and its higher frequencies.

Example 01:

A stretched wire of length one metre gives 3 beats per second with a tuning fork. If the length is made 0.99 m, no beats are heard between the two. What is the frequency of fork?

Given: Initial length = l₁ = 1m, Final length l₂ = 0.99 m, in case 1 |n₁ – N| = 3 beats per second, in case 2, n₂ = N

To Find: Frequency of Tuning fork = N =?

Solution:

By law length n α 1/l

l₁ > l₂, hence n₁ < n₂

n₂ – n₁ = 3

n₁ = n₂ – 3

By law of length

0.99 n₂ = n₂ – 3

3 = 0.01 n₂

n₂ = 3/0.01 = 300 Hz

Ans: Frequency of tuning fork is 300 Hz.

Example 02:

A sonometer wire one metre long is stretched by a certain tension and is divided into two segments by a movable knife edge so that the two segments of the vibrating wire differ in length by 2 mm. when set into vibration, the two segments produce 2 beats per second. Find their frequencies of vibration.

Given: l₁ – l₂ = 2 mm (assuming l₁ > l₂), |n₁ – n₂| = 2 beats per second.

To Find: Frequencies n₁ =? and n₂ =?

Solution:

l₁ + l₂ = 1m = 1000 mm

Thus l₁ = 501 mm and l₂ = 499 mm

By law length n α 1/l

l₁ > l₂, hence n₁ < n₂

n₂ – n₁ = 2

n₁ = n₂ – 2

By law of length

499 n₂ = 501 n₂ – 1002

1002 = 2 n₂

n₂ = 1002/2 = 501 Hz

n₁ = n₂ – 2 = 501 – 2 = 499 Hz

Ans: The frequencies of vibrations of two segments are 499 Hz and 501 Hz

Example 03:

A tuning fork of frequency 325 Hz is in unison with a sonometer wire of length 67 cm under certain tension. Find the number of beats heard per second when the length of the vibrating wire is decreased by 2 cm keeping the tension constant.

Given: l₁ = 67 cm, l₂ = 67 – 2 = 65 cm, Initial frequency = n₁ = N = 325 Hz

To Find: |n₁ – n₂| =?

Solution:

By law length n α 1/l

l₁ > l₂, hence n₁ < n₂

By law of length

Number of beats = n₂ – n₁ = 335 – 325 = 10 per second

Number of beats = 10 per second

Ans: Number of beats heard 10 per second i.e.  10 Hz

Example 04:

A sonometer wire 0.60 m long resonates to a tuning fork. When its length is decreased by 5 cm, 10 beats per second are heard. Find the frequency of fork. If the mass per unit length of wire is 0.014 g cm^-1, what is the tension?

Given: l₁ = 0.60 m = 60 cm, l₂ = 60 – 5 = 55 cm, |n₁ – n₂| = 10 per second, Mass per unit length = 0.014 g cm^-1 = 0.014 x 10^-1 kg m^-1 = 1.4 x 10^-3 kg m^-1,

To Find: Frequency of fork = n₁ = N =? Tension = T =?

Solution:

By law length n α 1/l

l₁ > l₂, hence n₁ < n₂

n₂ – n₁ = 10

n₂ = n₁ + 10

By law of length

55n₁ + 550 = 60 n₁

5n₁ = 550

n₁ = 110 Hz = N

Fundamental frequency of vibration of string is given by

Ans: Frequency of tuning fork 110 Hz, Tension in string is 24.39 N.

Example 05:

A sonometer wire under constant tension resonates with a tuning fork when the length of the wire is 65 cm. When the length of the wire is increased by one cm, 8 beats per second are heard with the same fork. Find the frequency of fork.

Given: l₁ = 65 cm, l₂ = 65 + 1 = 66 cm, |n₁ – n₂| = 8 per second.

To Find: Frequency of fork = n₁ = N =?

Solution:

By law length n α 1/l

l₁ < l₂, hence n₁ > n₂

n₁ – n₂ = 8

n₂ = n₁ – 8

By law of length

66n₁ – 528 = 65n₁

n₁ = 528 Hz

Ans: Frequency of tuning fork 528 Hz.

Example 06:

A stretched sonometer wire is in unison with a tuning fork. When the length of the wire is increased by two per cent, the number of beats heard per second is 5. Calculate the frequency of fork.

Given: Initial length = l₁ cm, Final length = l₂ = l₁ + 2% l₁ = 1.02 l₁, |n₁ – n₂| = 5 per second.

To Find: Frequency of fork = n₁ = N =?

Solution:

By law length n α 1/l

l₁ < l₂, hence n₁ > n₂

n₁ – n₂ = 5

n₂ = n₁ – 5

By law of length

1.02n₁ – 5.1 = n₁

0.02 n₁ = 5.1

N1 = 5.1/0.02 = 255 Hz

Ans: Frequency of tuning fork 255 Hz.

Example 07:

A length of sonometer wire under constant tension when sounded with a tuning fork of higher frequency 271 Hz produced 5 beats per second. If the length of the wire is reduced by 5%, find the new frequency of the wire.

Given: Initial length = l₁ cm, Final length = l₂ = l₁ – 5% l₁ = 0.95 l₁, n₁ = N – 5 = 271 – 5 = 266 Hz.

To Find: New frequency of wire = n₂ =?

Solution:

By law of length

Ans: New frequency of wire is 280 Hz.

Example 08:

When two tuning forks are sounded together, 5 beats per second are heard. One of the forks vibrates in unison with a sonometer wire of length 90 cm subjected to a given tension while the other fork vibrates with same wire of 91 cm length subjected to the same tension. Calculate the frequencies of tuning forks.

Given: Initial length = l₁ = 90 cm, Final length = l₂ = 91 cm, n₁ = N₁, n₂ = N₂, |N₁ – N₂| = 5 beats per second.

To Find: Frequencies of tuning forks, N₁ =? = N₂ =?

Solution:

By law length n α 1/l

l₁ < l₂, hence n₁ > n₂

N₁ > N₂

N₁ – N₂ = 5

N₁ = N₂ + 5

By law of length

91 N₂ = 90 N₂ + 450

N₂ = 450 Hz

N₁ = N₂ + 5 = 450 + 5 = 455 Hz

Ans: Frequencies of tuning forks are 455 Hz and 450 Hz.

Example 09:

Two tuning forks when sounded together give 5 beats per second. One is in unison with a sonometer wire of length 1.20 m under constant tension. The other tuning fork is in unison with the same wire of length 1.25 m under identical conditions. Find the frequencies of the two forks.

Given: Initial length = l₁ = 1.20 m, Final length = l₂ = 1.25 m, n₁ = N₁, n₂ = N₂, |N₁ – N₂| = 5 beats per second.

To Find: Frequencies of tuning forks, N₁ =? = N₂ =?

Solution:

By law length n α 1/l

l₁ < l₂, hence n₁ > n₂

N₁ > N₂

N₁ – N₂ = 5

N₁ = N₂ + 5

By law of length

1.25 N₂ = 1.20 N₂ + 6

0.05 N₂ = 6

N₂ = 6/0.05 = 120 Hz

N₁ = N₂ + 5 = 120 + 5 = 125 Hz

Ans: Frequencies of tuning forks are 125 Hz and 120 Hz.

Example 10:

Two tuning forks when sounded together produce 5 beats per second. A sonometer wire of length 0.24 m is in unison with one of the forks. When the length of wire is increased by 1 cm, it is in unison with other fork. What are the frequencies of the tuning forks?

Given: Initial length = l₁ = 0.24 m = 24 cm, Final length = l₂ = 24 + 1 = 25 cm, n₁ = N₁, n₂ = N₂, |N₁ – N₂| = 5 beats per second.

To Find: Frequencies of tuning forks, N₁ =? = N₂ =?

Solution:

By law length n α 1/l

l₁ < l₂, hence n₁ > n₂

N₁ > N₂

N₁ – N₂ = 5

N₁ = N₂ + 5

By law of length

25 N₂ = 24 N₂ + 120

N₂ = 120 Hz

N₁ = N₂ + 5 = 120 + 5 = 125 Hz

Ans: Frequencies of tuning forks are 125 Hz and 120 Hz.

Example 11:

A length of 34 cm of a sonometer wire gives 5 beats per second with a tuning fork. It again gives 5 beats per second when the length of the wire is increased to 35 cm. What is the frequency of fork?

Given: Initial length = l₁ = 34 cm, Final length = l₂ = 35 cm,

To Find: Frequency of tuning fork, N =?

Solution:

By law length n α 1/l

l₁ < l₂, hence n₁ > n₂

Now fork gives 5 beats per second in both cases

Hence, n₁ > N > n₂

n₁ – N = 5 i.e., n₁ = N + 5

N – n₂ = 5 i.e., n₂ = N – 5

By law of length

35N – 175 = 34N + 170

N = 345 Hz

Ans: Frequency of tuning fork is 345 Hz.

Example 12:

A tuning fork when sounded with a stretched sonometer wire produces 6 beats per second with the vibrating wire when its length is 95 cm or 100 cm. Find the frequency of the tuning fork.

Given: Initial length = l₁ = 95 cm, Final length = l₂ = 100 cm,

To Find: Frequency of tuning fork, N =?

Solution:

By law length n α 1/l

l₁ < l₂, hence n₁ > n₂

Now fork gives 5 beats per second in both cases

Hence, n₁ > N > n₂

n₁ – N = 6 i.e., n₁ = N + 6

N – n₂ = 6 i.e., n₂ = N – 6

By law of length

100N – 600 = 95N + 570

5N = 1170

N = 234 Hz

Ans: Frequency of tuning fork is 234 Hz.