172568
A stationary sound wave has a frequency of 165 Hz. If the speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$, then the distance between a node and the adjacent anti-node is
1 $20 \mathrm{~cm}$
2 $2 \mathrm{~m}$
3 $80 \mathrm{~cm}$
4 $50 \mathrm{~cm}$
Explanation:
D Given, Frequency $(\mathrm{f})=165 \mathrm{~Hz}$, speed of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ We know that, $\mathrm{v}=\mathrm{f} \lambda$ $\lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{330}{165}=2$ The distance between a node and the next anti-node $=$ $\frac{\lambda}{4}=\frac{2}{4}=\frac{1}{2}=0.5 \mathrm{~m}=50 \mathrm{~cm}$
MHT-CET 2020
WAVES
172569
A resonance air column of length $20 \mathrm{~cm}$ resonates with a tuning fork of frequency 250 Hz. The speed of sound in air is
1 $300 \mathrm{~m} / \mathrm{s}$
2 $200 \mathrm{~m} / \mathrm{s}$
3 $150 \mathrm{~m} / \mathrm{s}$
4 $75 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, The length of a resonance air column $(l)=20 \mathrm{~cm}$ Frequency $(\mathrm{f})=250 \mathrm{~Hz}$ $=0.2 \mathrm{~m}$ We know that, In resonance air column, $\lambda =4 l=4 \times 0.2=0.8 \mathrm{~m}$ $\because \quad \mathrm{V} =\mathrm{f} \lambda$ $\mathrm{V} =250 \times 0.8$ $\mathrm{~V} =200 \mathrm{~m} / \mathrm{s}$
UPSEE - 2010
WAVES
172570
Two open organ pipes of fundamental frequencies $n_{1}$ and $n_{2}$ are joined in series. The fundamental frequency of the new pipe is
C Fundamental frequency of an open pipe of length $l$ is given by, $\mathrm{n}=\frac{\mathrm{v}}{2 l}$ $\mathrm{n}_{1}=\frac{\mathrm{v}}{2 l_{1}} \Rightarrow l_{1}=\frac{\mathrm{v}}{2 \mathrm{n}_{1}}$ $\mathrm{n}_{2}=\frac{\mathrm{v}}{2 l_{2}} \Rightarrow l_{2}=\frac{\mathrm{v}}{2 \mathrm{n}_{2}}$ $l=l_{1}+l_{2}$ $\frac{\mathrm{v}}{2 \mathrm{n}}=\frac{\mathrm{v}}{2 \mathrm{n}_{1}}+\frac{\mathrm{v}}{2 \mathrm{n}_{2}}$ $\frac{1}{\mathrm{n}}=\frac{1}{\mathrm{n}_{1}}+\frac{1}{\mathrm{n}_{2}}=\frac{\mathrm{n}_{2}+\mathrm{n}_{1}}{\mathrm{n}_{1} \mathrm{n}_{2}}$ $\therefore \quad \mathrm{n} =\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
MHT-CET 2020
WAVES
172571
An open bamboo pipe has a fundamental frequency $(\mathrm{f})$. The tube is dipped vertically into water so that exactly half of the length is immersed. What will be the new fundamental frequency?
1 $f / 4$
2 $f / 2$
3 $\mathrm{f}$
4 $2 \mathrm{f}$
Explanation:
C For open ends of bamboo pipe fundamental frequency (f) in air: We know that, $\frac{\lambda}{2} =l$ $\lambda =2 l$ $\mathrm{v} =\mathrm{f} \lambda$ From equation (i) and (ii) we get - $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}$ When a bamboo pipe is dipped vertically in the water so that half of it is in water, then we have $\frac{\lambda}{4} =\frac{l}{2}$ $2 \lambda =4 l$ $\lambda =2 l$ $\therefore \quad \mathrm{f}^{\prime} =\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}=\mathrm{f}$ $\mathrm{f}^{\prime} =\mathrm{f}$
172568
A stationary sound wave has a frequency of 165 Hz. If the speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$, then the distance between a node and the adjacent anti-node is
1 $20 \mathrm{~cm}$
2 $2 \mathrm{~m}$
3 $80 \mathrm{~cm}$
4 $50 \mathrm{~cm}$
Explanation:
D Given, Frequency $(\mathrm{f})=165 \mathrm{~Hz}$, speed of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ We know that, $\mathrm{v}=\mathrm{f} \lambda$ $\lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{330}{165}=2$ The distance between a node and the next anti-node $=$ $\frac{\lambda}{4}=\frac{2}{4}=\frac{1}{2}=0.5 \mathrm{~m}=50 \mathrm{~cm}$
MHT-CET 2020
WAVES
172569
A resonance air column of length $20 \mathrm{~cm}$ resonates with a tuning fork of frequency 250 Hz. The speed of sound in air is
1 $300 \mathrm{~m} / \mathrm{s}$
2 $200 \mathrm{~m} / \mathrm{s}$
3 $150 \mathrm{~m} / \mathrm{s}$
4 $75 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, The length of a resonance air column $(l)=20 \mathrm{~cm}$ Frequency $(\mathrm{f})=250 \mathrm{~Hz}$ $=0.2 \mathrm{~m}$ We know that, In resonance air column, $\lambda =4 l=4 \times 0.2=0.8 \mathrm{~m}$ $\because \quad \mathrm{V} =\mathrm{f} \lambda$ $\mathrm{V} =250 \times 0.8$ $\mathrm{~V} =200 \mathrm{~m} / \mathrm{s}$
UPSEE - 2010
WAVES
172570
Two open organ pipes of fundamental frequencies $n_{1}$ and $n_{2}$ are joined in series. The fundamental frequency of the new pipe is
C Fundamental frequency of an open pipe of length $l$ is given by, $\mathrm{n}=\frac{\mathrm{v}}{2 l}$ $\mathrm{n}_{1}=\frac{\mathrm{v}}{2 l_{1}} \Rightarrow l_{1}=\frac{\mathrm{v}}{2 \mathrm{n}_{1}}$ $\mathrm{n}_{2}=\frac{\mathrm{v}}{2 l_{2}} \Rightarrow l_{2}=\frac{\mathrm{v}}{2 \mathrm{n}_{2}}$ $l=l_{1}+l_{2}$ $\frac{\mathrm{v}}{2 \mathrm{n}}=\frac{\mathrm{v}}{2 \mathrm{n}_{1}}+\frac{\mathrm{v}}{2 \mathrm{n}_{2}}$ $\frac{1}{\mathrm{n}}=\frac{1}{\mathrm{n}_{1}}+\frac{1}{\mathrm{n}_{2}}=\frac{\mathrm{n}_{2}+\mathrm{n}_{1}}{\mathrm{n}_{1} \mathrm{n}_{2}}$ $\therefore \quad \mathrm{n} =\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
MHT-CET 2020
WAVES
172571
An open bamboo pipe has a fundamental frequency $(\mathrm{f})$. The tube is dipped vertically into water so that exactly half of the length is immersed. What will be the new fundamental frequency?
1 $f / 4$
2 $f / 2$
3 $\mathrm{f}$
4 $2 \mathrm{f}$
Explanation:
C For open ends of bamboo pipe fundamental frequency (f) in air: We know that, $\frac{\lambda}{2} =l$ $\lambda =2 l$ $\mathrm{v} =\mathrm{f} \lambda$ From equation (i) and (ii) we get - $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}$ When a bamboo pipe is dipped vertically in the water so that half of it is in water, then we have $\frac{\lambda}{4} =\frac{l}{2}$ $2 \lambda =4 l$ $\lambda =2 l$ $\therefore \quad \mathrm{f}^{\prime} =\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}=\mathrm{f}$ $\mathrm{f}^{\prime} =\mathrm{f}$
172568
A stationary sound wave has a frequency of 165 Hz. If the speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$, then the distance between a node and the adjacent anti-node is
1 $20 \mathrm{~cm}$
2 $2 \mathrm{~m}$
3 $80 \mathrm{~cm}$
4 $50 \mathrm{~cm}$
Explanation:
D Given, Frequency $(\mathrm{f})=165 \mathrm{~Hz}$, speed of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ We know that, $\mathrm{v}=\mathrm{f} \lambda$ $\lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{330}{165}=2$ The distance between a node and the next anti-node $=$ $\frac{\lambda}{4}=\frac{2}{4}=\frac{1}{2}=0.5 \mathrm{~m}=50 \mathrm{~cm}$
MHT-CET 2020
WAVES
172569
A resonance air column of length $20 \mathrm{~cm}$ resonates with a tuning fork of frequency 250 Hz. The speed of sound in air is
1 $300 \mathrm{~m} / \mathrm{s}$
2 $200 \mathrm{~m} / \mathrm{s}$
3 $150 \mathrm{~m} / \mathrm{s}$
4 $75 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, The length of a resonance air column $(l)=20 \mathrm{~cm}$ Frequency $(\mathrm{f})=250 \mathrm{~Hz}$ $=0.2 \mathrm{~m}$ We know that, In resonance air column, $\lambda =4 l=4 \times 0.2=0.8 \mathrm{~m}$ $\because \quad \mathrm{V} =\mathrm{f} \lambda$ $\mathrm{V} =250 \times 0.8$ $\mathrm{~V} =200 \mathrm{~m} / \mathrm{s}$
UPSEE - 2010
WAVES
172570
Two open organ pipes of fundamental frequencies $n_{1}$ and $n_{2}$ are joined in series. The fundamental frequency of the new pipe is
C Fundamental frequency of an open pipe of length $l$ is given by, $\mathrm{n}=\frac{\mathrm{v}}{2 l}$ $\mathrm{n}_{1}=\frac{\mathrm{v}}{2 l_{1}} \Rightarrow l_{1}=\frac{\mathrm{v}}{2 \mathrm{n}_{1}}$ $\mathrm{n}_{2}=\frac{\mathrm{v}}{2 l_{2}} \Rightarrow l_{2}=\frac{\mathrm{v}}{2 \mathrm{n}_{2}}$ $l=l_{1}+l_{2}$ $\frac{\mathrm{v}}{2 \mathrm{n}}=\frac{\mathrm{v}}{2 \mathrm{n}_{1}}+\frac{\mathrm{v}}{2 \mathrm{n}_{2}}$ $\frac{1}{\mathrm{n}}=\frac{1}{\mathrm{n}_{1}}+\frac{1}{\mathrm{n}_{2}}=\frac{\mathrm{n}_{2}+\mathrm{n}_{1}}{\mathrm{n}_{1} \mathrm{n}_{2}}$ $\therefore \quad \mathrm{n} =\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
MHT-CET 2020
WAVES
172571
An open bamboo pipe has a fundamental frequency $(\mathrm{f})$. The tube is dipped vertically into water so that exactly half of the length is immersed. What will be the new fundamental frequency?
1 $f / 4$
2 $f / 2$
3 $\mathrm{f}$
4 $2 \mathrm{f}$
Explanation:
C For open ends of bamboo pipe fundamental frequency (f) in air: We know that, $\frac{\lambda}{2} =l$ $\lambda =2 l$ $\mathrm{v} =\mathrm{f} \lambda$ From equation (i) and (ii) we get - $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}$ When a bamboo pipe is dipped vertically in the water so that half of it is in water, then we have $\frac{\lambda}{4} =\frac{l}{2}$ $2 \lambda =4 l$ $\lambda =2 l$ $\therefore \quad \mathrm{f}^{\prime} =\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}=\mathrm{f}$ $\mathrm{f}^{\prime} =\mathrm{f}$
172568
A stationary sound wave has a frequency of 165 Hz. If the speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$, then the distance between a node and the adjacent anti-node is
1 $20 \mathrm{~cm}$
2 $2 \mathrm{~m}$
3 $80 \mathrm{~cm}$
4 $50 \mathrm{~cm}$
Explanation:
D Given, Frequency $(\mathrm{f})=165 \mathrm{~Hz}$, speed of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ We know that, $\mathrm{v}=\mathrm{f} \lambda$ $\lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{330}{165}=2$ The distance between a node and the next anti-node $=$ $\frac{\lambda}{4}=\frac{2}{4}=\frac{1}{2}=0.5 \mathrm{~m}=50 \mathrm{~cm}$
MHT-CET 2020
WAVES
172569
A resonance air column of length $20 \mathrm{~cm}$ resonates with a tuning fork of frequency 250 Hz. The speed of sound in air is
1 $300 \mathrm{~m} / \mathrm{s}$
2 $200 \mathrm{~m} / \mathrm{s}$
3 $150 \mathrm{~m} / \mathrm{s}$
4 $75 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, The length of a resonance air column $(l)=20 \mathrm{~cm}$ Frequency $(\mathrm{f})=250 \mathrm{~Hz}$ $=0.2 \mathrm{~m}$ We know that, In resonance air column, $\lambda =4 l=4 \times 0.2=0.8 \mathrm{~m}$ $\because \quad \mathrm{V} =\mathrm{f} \lambda$ $\mathrm{V} =250 \times 0.8$ $\mathrm{~V} =200 \mathrm{~m} / \mathrm{s}$
UPSEE - 2010
WAVES
172570
Two open organ pipes of fundamental frequencies $n_{1}$ and $n_{2}$ are joined in series. The fundamental frequency of the new pipe is
C Fundamental frequency of an open pipe of length $l$ is given by, $\mathrm{n}=\frac{\mathrm{v}}{2 l}$ $\mathrm{n}_{1}=\frac{\mathrm{v}}{2 l_{1}} \Rightarrow l_{1}=\frac{\mathrm{v}}{2 \mathrm{n}_{1}}$ $\mathrm{n}_{2}=\frac{\mathrm{v}}{2 l_{2}} \Rightarrow l_{2}=\frac{\mathrm{v}}{2 \mathrm{n}_{2}}$ $l=l_{1}+l_{2}$ $\frac{\mathrm{v}}{2 \mathrm{n}}=\frac{\mathrm{v}}{2 \mathrm{n}_{1}}+\frac{\mathrm{v}}{2 \mathrm{n}_{2}}$ $\frac{1}{\mathrm{n}}=\frac{1}{\mathrm{n}_{1}}+\frac{1}{\mathrm{n}_{2}}=\frac{\mathrm{n}_{2}+\mathrm{n}_{1}}{\mathrm{n}_{1} \mathrm{n}_{2}}$ $\therefore \quad \mathrm{n} =\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
MHT-CET 2020
WAVES
172571
An open bamboo pipe has a fundamental frequency $(\mathrm{f})$. The tube is dipped vertically into water so that exactly half of the length is immersed. What will be the new fundamental frequency?
1 $f / 4$
2 $f / 2$
3 $\mathrm{f}$
4 $2 \mathrm{f}$
Explanation:
C For open ends of bamboo pipe fundamental frequency (f) in air: We know that, $\frac{\lambda}{2} =l$ $\lambda =2 l$ $\mathrm{v} =\mathrm{f} \lambda$ From equation (i) and (ii) we get - $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}$ When a bamboo pipe is dipped vertically in the water so that half of it is in water, then we have $\frac{\lambda}{4} =\frac{l}{2}$ $2 \lambda =4 l$ $\lambda =2 l$ $\therefore \quad \mathrm{f}^{\prime} =\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}=\mathrm{f}$ $\mathrm{f}^{\prime} =\mathrm{f}$
