172419
A transverse wave $Y=2 \sin (0.01 x+30 t)$ moves on a stretched string from one end to another end in $0.5 \mathrm{~s}$. If $x$ and $Y$ are in $\mathrm{cm}$ and $t$ in second, then the length of the string is
1 $20 \mathrm{~m}$
2 $15 \mathrm{~m}$
3 $10 \mathrm{~m}$
4 $5 \mathrm{~m}$
Explanation:
B Transverse wave $\mathrm{Y}=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})$ Standard wave equation $\mathrm{Y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})$ On comparing both the equation- $\mathrm{k}=0.01 \mathrm{~cm}^{-1}$ $\omega=30 \mathrm{rad} / \mathrm{sec}$ Wave moves from one end to another end in $0.5 \mathrm{sec}$. Speed of wave $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=3000 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}$ So, length of the string $(\mathrm{L})=\mathrm{v} \times \mathrm{t}=30 \times 0.5=15 \mathrm{~m}$
MHT-CET 2020
WAVES
172420
The fundamental frequency of a string stretched with a weight ' $M$ ' $\mathrm{kg}$ is ' $\mathrm{n}$ ' hertz. Keeping the vibrating length constant, the weight required to produce its octave is
1 $\mathrm{M}$
2 $4 \mathrm{M}$
3 $2 \mathrm{M}$
4 $8 \mathrm{M}$
Explanation:
B Length of string $=l$ Linear mass density of string $=\mu$ Fundamental frequency $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ Octave means that the frequency is in the ratio $\mathrm{n}_{1}: \mathrm{n}_{2}=2: 1$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}$ $\mathrm{n}_{1}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{1}}{\mu}}$ $\mathrm{n}_{2}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{2}}{\mu}}$ Dividing equation (i) by equation (ii), we get $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{\mathrm{n}_{1}}{2 \mathrm{n}_{1}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ Squaring both sides- $\left(\frac{1}{2}\right)^{2} =\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =4 \mathrm{~T}_{1} \quad\left[\because \mathrm{T}_{1}=\mathrm{M}\right]$ $\mathrm{T}_{2} =4 \mathrm{M}$
MHT-CET 2020
WAVES
172421
A string of length ' $L$ ' and linear mass density ' $\mu$ ' has a fundamental frequency ' $n$ ' when stretched by tension ' $T$ '. The fundamental frequency of another string having double the length and double linear density, when same tension is applied is
1 $\frac{\mathrm{n}}{2}$
2 $\frac{\mathrm{n}}{2 \sqrt{2}}$
3 $\frac{\mathrm{n}}{\sqrt{2}}$
4 $2 \mathrm{n}$
Explanation:
B Given, Length of string $=\mathrm{L}$ Linear mass density $=\mu$ Fundamental frequency $=\mathrm{n}$ Tension $=\mathrm{T}$ $\because \mathrm{n}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$ Fundamental frequency of another string having Length $=2 \mathrm{~L}$ Linear mass density $=2 \mu$ Tension $=\mathrm{T}$ $\therefore \mathrm{n}^{\prime}=\frac{1}{2 \times(2 \mathrm{~L})} \sqrt{\frac{\mathrm{T}}{2 \mu}}$ Dividing equation (i) by (ii), we get- $\frac{\mathrm{n}}{\mathrm{n}^{\prime}} =\frac{2 \sqrt{2}}{1}$ $\mathrm{n}^{\prime} =\frac{\mathrm{n}}{2 \sqrt{2}}$
MHT-CET 2020
WAVES
172422
The equation of stationary wave on a string clamped at both ends and vibrating in third harmonic is given by $y=0.5 \sin (0.314 x) \cos$ $(600 \pi t)$, where $x$ and $y$ are in $\mathrm{cm}$ and $t$ in second. The length of the vibrating string is $(\pi$ =3.14)
1 $15 \mathrm{~cm}$
2 $40 \mathrm{~cm}$
3 $10 \mathrm{~cm}$
4 $30 \mathrm{~cm}$
Explanation:
D Stationary wave $y=0.5 \sin (0.314 x) \cos (600 \pi t)$ Standard stationary wave equation $\mathrm{y}=2 \mathrm{~A} \sin \left(\frac{2 \pi}{\lambda} \mathrm{x}\right) \cos (2 \pi \mathrm{nt})$ Comparing both the equation, we get - $\frac{2 \pi}{\lambda}=0.314$ $\lambda=\frac{2 \times 3.14}{0.314}$ $\lambda=\frac{2 \times 3.14 \times 10}{3.14}=20 \mathrm{~cm}$ String is vibrating in third harmonic, forming three loops so, length of string is $\mathrm{L}=3 \frac{\lambda}{2}$ $\mathrm{~L}=3 \times \frac{20}{2}=30 \mathrm{~cm}$
172419
A transverse wave $Y=2 \sin (0.01 x+30 t)$ moves on a stretched string from one end to another end in $0.5 \mathrm{~s}$. If $x$ and $Y$ are in $\mathrm{cm}$ and $t$ in second, then the length of the string is
1 $20 \mathrm{~m}$
2 $15 \mathrm{~m}$
3 $10 \mathrm{~m}$
4 $5 \mathrm{~m}$
Explanation:
B Transverse wave $\mathrm{Y}=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})$ Standard wave equation $\mathrm{Y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})$ On comparing both the equation- $\mathrm{k}=0.01 \mathrm{~cm}^{-1}$ $\omega=30 \mathrm{rad} / \mathrm{sec}$ Wave moves from one end to another end in $0.5 \mathrm{sec}$. Speed of wave $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=3000 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}$ So, length of the string $(\mathrm{L})=\mathrm{v} \times \mathrm{t}=30 \times 0.5=15 \mathrm{~m}$
MHT-CET 2020
WAVES
172420
The fundamental frequency of a string stretched with a weight ' $M$ ' $\mathrm{kg}$ is ' $\mathrm{n}$ ' hertz. Keeping the vibrating length constant, the weight required to produce its octave is
1 $\mathrm{M}$
2 $4 \mathrm{M}$
3 $2 \mathrm{M}$
4 $8 \mathrm{M}$
Explanation:
B Length of string $=l$ Linear mass density of string $=\mu$ Fundamental frequency $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ Octave means that the frequency is in the ratio $\mathrm{n}_{1}: \mathrm{n}_{2}=2: 1$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}$ $\mathrm{n}_{1}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{1}}{\mu}}$ $\mathrm{n}_{2}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{2}}{\mu}}$ Dividing equation (i) by equation (ii), we get $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{\mathrm{n}_{1}}{2 \mathrm{n}_{1}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ Squaring both sides- $\left(\frac{1}{2}\right)^{2} =\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =4 \mathrm{~T}_{1} \quad\left[\because \mathrm{T}_{1}=\mathrm{M}\right]$ $\mathrm{T}_{2} =4 \mathrm{M}$
MHT-CET 2020
WAVES
172421
A string of length ' $L$ ' and linear mass density ' $\mu$ ' has a fundamental frequency ' $n$ ' when stretched by tension ' $T$ '. The fundamental frequency of another string having double the length and double linear density, when same tension is applied is
1 $\frac{\mathrm{n}}{2}$
2 $\frac{\mathrm{n}}{2 \sqrt{2}}$
3 $\frac{\mathrm{n}}{\sqrt{2}}$
4 $2 \mathrm{n}$
Explanation:
B Given, Length of string $=\mathrm{L}$ Linear mass density $=\mu$ Fundamental frequency $=\mathrm{n}$ Tension $=\mathrm{T}$ $\because \mathrm{n}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$ Fundamental frequency of another string having Length $=2 \mathrm{~L}$ Linear mass density $=2 \mu$ Tension $=\mathrm{T}$ $\therefore \mathrm{n}^{\prime}=\frac{1}{2 \times(2 \mathrm{~L})} \sqrt{\frac{\mathrm{T}}{2 \mu}}$ Dividing equation (i) by (ii), we get- $\frac{\mathrm{n}}{\mathrm{n}^{\prime}} =\frac{2 \sqrt{2}}{1}$ $\mathrm{n}^{\prime} =\frac{\mathrm{n}}{2 \sqrt{2}}$
MHT-CET 2020
WAVES
172422
The equation of stationary wave on a string clamped at both ends and vibrating in third harmonic is given by $y=0.5 \sin (0.314 x) \cos$ $(600 \pi t)$, where $x$ and $y$ are in $\mathrm{cm}$ and $t$ in second. The length of the vibrating string is $(\pi$ =3.14)
1 $15 \mathrm{~cm}$
2 $40 \mathrm{~cm}$
3 $10 \mathrm{~cm}$
4 $30 \mathrm{~cm}$
Explanation:
D Stationary wave $y=0.5 \sin (0.314 x) \cos (600 \pi t)$ Standard stationary wave equation $\mathrm{y}=2 \mathrm{~A} \sin \left(\frac{2 \pi}{\lambda} \mathrm{x}\right) \cos (2 \pi \mathrm{nt})$ Comparing both the equation, we get - $\frac{2 \pi}{\lambda}=0.314$ $\lambda=\frac{2 \times 3.14}{0.314}$ $\lambda=\frac{2 \times 3.14 \times 10}{3.14}=20 \mathrm{~cm}$ String is vibrating in third harmonic, forming three loops so, length of string is $\mathrm{L}=3 \frac{\lambda}{2}$ $\mathrm{~L}=3 \times \frac{20}{2}=30 \mathrm{~cm}$
172419
A transverse wave $Y=2 \sin (0.01 x+30 t)$ moves on a stretched string from one end to another end in $0.5 \mathrm{~s}$. If $x$ and $Y$ are in $\mathrm{cm}$ and $t$ in second, then the length of the string is
1 $20 \mathrm{~m}$
2 $15 \mathrm{~m}$
3 $10 \mathrm{~m}$
4 $5 \mathrm{~m}$
Explanation:
B Transverse wave $\mathrm{Y}=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})$ Standard wave equation $\mathrm{Y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})$ On comparing both the equation- $\mathrm{k}=0.01 \mathrm{~cm}^{-1}$ $\omega=30 \mathrm{rad} / \mathrm{sec}$ Wave moves from one end to another end in $0.5 \mathrm{sec}$. Speed of wave $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=3000 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}$ So, length of the string $(\mathrm{L})=\mathrm{v} \times \mathrm{t}=30 \times 0.5=15 \mathrm{~m}$
MHT-CET 2020
WAVES
172420
The fundamental frequency of a string stretched with a weight ' $M$ ' $\mathrm{kg}$ is ' $\mathrm{n}$ ' hertz. Keeping the vibrating length constant, the weight required to produce its octave is
1 $\mathrm{M}$
2 $4 \mathrm{M}$
3 $2 \mathrm{M}$
4 $8 \mathrm{M}$
Explanation:
B Length of string $=l$ Linear mass density of string $=\mu$ Fundamental frequency $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ Octave means that the frequency is in the ratio $\mathrm{n}_{1}: \mathrm{n}_{2}=2: 1$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}$ $\mathrm{n}_{1}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{1}}{\mu}}$ $\mathrm{n}_{2}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{2}}{\mu}}$ Dividing equation (i) by equation (ii), we get $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{\mathrm{n}_{1}}{2 \mathrm{n}_{1}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ Squaring both sides- $\left(\frac{1}{2}\right)^{2} =\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =4 \mathrm{~T}_{1} \quad\left[\because \mathrm{T}_{1}=\mathrm{M}\right]$ $\mathrm{T}_{2} =4 \mathrm{M}$
MHT-CET 2020
WAVES
172421
A string of length ' $L$ ' and linear mass density ' $\mu$ ' has a fundamental frequency ' $n$ ' when stretched by tension ' $T$ '. The fundamental frequency of another string having double the length and double linear density, when same tension is applied is
1 $\frac{\mathrm{n}}{2}$
2 $\frac{\mathrm{n}}{2 \sqrt{2}}$
3 $\frac{\mathrm{n}}{\sqrt{2}}$
4 $2 \mathrm{n}$
Explanation:
B Given, Length of string $=\mathrm{L}$ Linear mass density $=\mu$ Fundamental frequency $=\mathrm{n}$ Tension $=\mathrm{T}$ $\because \mathrm{n}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$ Fundamental frequency of another string having Length $=2 \mathrm{~L}$ Linear mass density $=2 \mu$ Tension $=\mathrm{T}$ $\therefore \mathrm{n}^{\prime}=\frac{1}{2 \times(2 \mathrm{~L})} \sqrt{\frac{\mathrm{T}}{2 \mu}}$ Dividing equation (i) by (ii), we get- $\frac{\mathrm{n}}{\mathrm{n}^{\prime}} =\frac{2 \sqrt{2}}{1}$ $\mathrm{n}^{\prime} =\frac{\mathrm{n}}{2 \sqrt{2}}$
MHT-CET 2020
WAVES
172422
The equation of stationary wave on a string clamped at both ends and vibrating in third harmonic is given by $y=0.5 \sin (0.314 x) \cos$ $(600 \pi t)$, where $x$ and $y$ are in $\mathrm{cm}$ and $t$ in second. The length of the vibrating string is $(\pi$ =3.14)
1 $15 \mathrm{~cm}$
2 $40 \mathrm{~cm}$
3 $10 \mathrm{~cm}$
4 $30 \mathrm{~cm}$
Explanation:
D Stationary wave $y=0.5 \sin (0.314 x) \cos (600 \pi t)$ Standard stationary wave equation $\mathrm{y}=2 \mathrm{~A} \sin \left(\frac{2 \pi}{\lambda} \mathrm{x}\right) \cos (2 \pi \mathrm{nt})$ Comparing both the equation, we get - $\frac{2 \pi}{\lambda}=0.314$ $\lambda=\frac{2 \times 3.14}{0.314}$ $\lambda=\frac{2 \times 3.14 \times 10}{3.14}=20 \mathrm{~cm}$ String is vibrating in third harmonic, forming three loops so, length of string is $\mathrm{L}=3 \frac{\lambda}{2}$ $\mathrm{~L}=3 \times \frac{20}{2}=30 \mathrm{~cm}$
172419
A transverse wave $Y=2 \sin (0.01 x+30 t)$ moves on a stretched string from one end to another end in $0.5 \mathrm{~s}$. If $x$ and $Y$ are in $\mathrm{cm}$ and $t$ in second, then the length of the string is
1 $20 \mathrm{~m}$
2 $15 \mathrm{~m}$
3 $10 \mathrm{~m}$
4 $5 \mathrm{~m}$
Explanation:
B Transverse wave $\mathrm{Y}=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})$ Standard wave equation $\mathrm{Y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})$ On comparing both the equation- $\mathrm{k}=0.01 \mathrm{~cm}^{-1}$ $\omega=30 \mathrm{rad} / \mathrm{sec}$ Wave moves from one end to another end in $0.5 \mathrm{sec}$. Speed of wave $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=3000 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}$ So, length of the string $(\mathrm{L})=\mathrm{v} \times \mathrm{t}=30 \times 0.5=15 \mathrm{~m}$
MHT-CET 2020
WAVES
172420
The fundamental frequency of a string stretched with a weight ' $M$ ' $\mathrm{kg}$ is ' $\mathrm{n}$ ' hertz. Keeping the vibrating length constant, the weight required to produce its octave is
1 $\mathrm{M}$
2 $4 \mathrm{M}$
3 $2 \mathrm{M}$
4 $8 \mathrm{M}$
Explanation:
B Length of string $=l$ Linear mass density of string $=\mu$ Fundamental frequency $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$ Octave means that the frequency is in the ratio $\mathrm{n}_{1}: \mathrm{n}_{2}=2: 1$ $\mathrm{n}_{2}=2 \mathrm{n}_{1}$ $\mathrm{n}_{1}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{1}}{\mu}}$ $\mathrm{n}_{2}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}_{2}}{\mu}}$ Dividing equation (i) by equation (ii), we get $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ $\frac{\mathrm{n}_{1}}{2 \mathrm{n}_{1}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$ Squaring both sides- $\left(\frac{1}{2}\right)^{2} =\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\mathrm{~T}_{2} =4 \mathrm{~T}_{1} \quad\left[\because \mathrm{T}_{1}=\mathrm{M}\right]$ $\mathrm{T}_{2} =4 \mathrm{M}$
MHT-CET 2020
WAVES
172421
A string of length ' $L$ ' and linear mass density ' $\mu$ ' has a fundamental frequency ' $n$ ' when stretched by tension ' $T$ '. The fundamental frequency of another string having double the length and double linear density, when same tension is applied is
1 $\frac{\mathrm{n}}{2}$
2 $\frac{\mathrm{n}}{2 \sqrt{2}}$
3 $\frac{\mathrm{n}}{\sqrt{2}}$
4 $2 \mathrm{n}$
Explanation:
B Given, Length of string $=\mathrm{L}$ Linear mass density $=\mu$ Fundamental frequency $=\mathrm{n}$ Tension $=\mathrm{T}$ $\because \mathrm{n}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$ Fundamental frequency of another string having Length $=2 \mathrm{~L}$ Linear mass density $=2 \mu$ Tension $=\mathrm{T}$ $\therefore \mathrm{n}^{\prime}=\frac{1}{2 \times(2 \mathrm{~L})} \sqrt{\frac{\mathrm{T}}{2 \mu}}$ Dividing equation (i) by (ii), we get- $\frac{\mathrm{n}}{\mathrm{n}^{\prime}} =\frac{2 \sqrt{2}}{1}$ $\mathrm{n}^{\prime} =\frac{\mathrm{n}}{2 \sqrt{2}}$
MHT-CET 2020
WAVES
172422
The equation of stationary wave on a string clamped at both ends and vibrating in third harmonic is given by $y=0.5 \sin (0.314 x) \cos$ $(600 \pi t)$, where $x$ and $y$ are in $\mathrm{cm}$ and $t$ in second. The length of the vibrating string is $(\pi$ =3.14)
1 $15 \mathrm{~cm}$
2 $40 \mathrm{~cm}$
3 $10 \mathrm{~cm}$
4 $30 \mathrm{~cm}$
Explanation:
D Stationary wave $y=0.5 \sin (0.314 x) \cos (600 \pi t)$ Standard stationary wave equation $\mathrm{y}=2 \mathrm{~A} \sin \left(\frac{2 \pi}{\lambda} \mathrm{x}\right) \cos (2 \pi \mathrm{nt})$ Comparing both the equation, we get - $\frac{2 \pi}{\lambda}=0.314$ $\lambda=\frac{2 \times 3.14}{0.314}$ $\lambda=\frac{2 \times 3.14 \times 10}{3.14}=20 \mathrm{~cm}$ String is vibrating in third harmonic, forming three loops so, length of string is $\mathrm{L}=3 \frac{\lambda}{2}$ $\mathrm{~L}=3 \times \frac{20}{2}=30 \mathrm{~cm}$
