A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is:
\[
(a){\text{ 105Hz}} \\
(b){\text{ 155Hz}} \\
(c){\text{ 205Hz}} \\
(d){\text{ 10}}{\text{.5Hz}} \\
\]
Answer
667.8k+ views
Hint: In this question use the direct formula for frequency that is $f = \dfrac{v}{{2L}}n$ where v is velocity of sound, n is the smallest integer and L is the separation between two points of the strings. Use the concept that the highest common factor (H.C.F) is calculated by multiplying all the common factors of the given numbers. This will help reach the answer.
Complete step-by-step answer:
Given data:
String is stretched between fixed point separated by the distance = 75 cm
Therefore, L = 75cm
It is observed that the resonant frequencies of 420 Hz and 315 Hz.
There are no other resonant frequencies present between these two points.
Now we have to calculate the lowest resonant frequency for this string.
As we know that the frequency is given as $f = \dfrac{v}{{2L}}n$
Where, n = smallest integer
v = velocity of sound
L = separation length of the string by which it is stretched.
Now the lowest possible resonant frequency is the highest common factor (H.C.F) of the two given resonant frequencies.
As we know that the highest common factor (H.C.F) is calculated by multiplying all the common factors of the given numbers.
So first factories the numbers,
Therefore the factors of 420 are,
$420 = 2 \times 2 \times 3 \times 5 \times 7$
The factors of 105 are,
$315 = 3 \times 3 \times 5 \times 7$
Now, as we see that the common factors of above numbers are $3 \times 5 \times 7$.
So the required H.C.F of the given numbers is $3 \times 5 \times 7 = 105$.
So, H.C.F of 420 and 315 is 105
So the lowest resonant frequency for this string is 105 Hz.
So this is the required answer.
Hence option (A) is the correct answer.
Note – There are basically two types of oscillations : one is natural and one is forced. When a body starts oscillating at its natural then this frequency is termed as the resonant frequency. Resonance occurs when the frequency of the body undergoing an external force matches up to its natural oscillating frequency.
Complete step-by-step answer:
Given data:
String is stretched between fixed point separated by the distance = 75 cm
Therefore, L = 75cm
It is observed that the resonant frequencies of 420 Hz and 315 Hz.
There are no other resonant frequencies present between these two points.
Now we have to calculate the lowest resonant frequency for this string.
As we know that the frequency is given as $f = \dfrac{v}{{2L}}n$
Where, n = smallest integer
v = velocity of sound
L = separation length of the string by which it is stretched.
Now the lowest possible resonant frequency is the highest common factor (H.C.F) of the two given resonant frequencies.
As we know that the highest common factor (H.C.F) is calculated by multiplying all the common factors of the given numbers.
So first factories the numbers,
Therefore the factors of 420 are,
$420 = 2 \times 2 \times 3 \times 5 \times 7$
The factors of 105 are,
$315 = 3 \times 3 \times 5 \times 7$
Now, as we see that the common factors of above numbers are $3 \times 5 \times 7$.
So the required H.C.F of the given numbers is $3 \times 5 \times 7 = 105$.
So, H.C.F of 420 and 315 is 105
So the lowest resonant frequency for this string is 105 Hz.
So this is the required answer.
Hence option (A) is the correct answer.
Note – There are basically two types of oscillations : one is natural and one is forced. When a body starts oscillating at its natural then this frequency is termed as the resonant frequency. Resonance occurs when the frequency of the body undergoing an external force matches up to its natural oscillating frequency.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

