355182
The speed of transverse waves in a stretched string is \({700 {~cm} / {s}}\). It the string is \(2\,m\) long, the frequency with which it resonates in fundamental mode is
1 \({7 / 2 {~Hz}}\)
2 \({7 / 4 {~Hz}}\)
3 14 Hz
4 \({2 / 7 {~Hz}}\)
Explanation:
\({v=700 {~cm} / {s}=7 {~m} / {s}}\) For one loop, \({l=\lambda / 2}\) \({\lambda=2 l=2 \times 2=4 m}\) \({f=\dfrac{v}{\lambda}=7 / 4 H z}\)
PHXI15:WAVES
355183
The expression \(y=a \sin b x \sin \omega t\) represents a stationary wave. The distance between the consecutive nodes is equal to:
1 \(2 \pi / b\)
2 \(\pi / b\)
3 \(1 / b\)
4 \(\pi / 2 b\)
Explanation:
\(y=a \sin b x \sin \omega t\) On comparing with standard equation of stationary wave \(\begin{aligned}& y=2 a \sin \dfrac{2 \pi x}{\lambda} \cdot \sin \omega t, \text { we get } \\& \dfrac{2 \pi x}{\lambda}=b x, \\& \therefore \lambda=\dfrac{2 \pi}{b}\end{aligned}\) The distance between constructive nodes \(=\dfrac{\lambda}{2}=\dfrac{\pi}{b}\)
PHXI15:WAVES
355184
A string of mass \(m\) is fixed at both ends. The Fundamental mode of string is exicted and it has an angular frequency \(\omega\) and the maximum displacement amplitude A.then \(K.E\) for one periodic time is
1 \(\dfrac{1}{2} m A^{2} \omega^{2}\)
2 \(m A^{2} \omega^{2}\)
3 \(\dfrac{1}{4} m A^{2} \omega^{2}\)
4 \(\dfrac{1}{8} m A^{2} \omega^{2}\)
Explanation:
Let \(y=A \sin k x \sin \omega t\) represents the standing wave fixed at both ends. The velocity of a particle is \(v_{p}=\dfrac{\partial y}{\partial t}\) \(v_{p}=A \omega \sin k x \cos \omega t\) The small \(K.E\) of the element is \(\begin{aligned}& d k=\dfrac{1}{2} d m v_{p}^{2}=\dfrac{1}{2} \mu d x v_{p}^{2} \\& d k=\dfrac{1}{2} \mu A^{2} \omega^{2} \sin ^{2} k x \cos ^{2} \omega t d x \\& k=\dfrac{1}{2} \mu A^{2} \omega^{2} \cos ^{2} \omega t \int_{0}^{L} \sin ^{2} k x d x \\& k=\dfrac{1}{4} \mu A^{2} \omega^{2} \cos ^{2} \omega t \int_{0}^{L}(1-\cos 2 k x) d x \\& k=\dfrac{1}{4} \mu A^{2} \omega^{2} \cos ^{2} \omega t\left[L-\dfrac{1}{2 k}[\sin 2 k x]_{0}^{L}\right]\end{aligned}\) given that the string is in Fundamental mode \(\begin{aligned}& \dfrac{\lambda}{2}=L \Rightarrow \lambda=2 L \Rightarrow k=\dfrac{2 \pi}{2 L} \\& k=\dfrac{1}{4} \mu L A^{2} \omega^{2} \cos ^{2} \omega t \\& =\dfrac{1}{4} m A^{2} \omega^{2} \cos ^{2} \omega t \quad(\mu L=m) \\& < k>=\dfrac{1}{4} m A^{2} \omega^{2} < \cos ^{2} \omega t>=\dfrac{1}{8} m A^{2} \omega^{2}\end{aligned}\) (for one time period) \(\left[ < \sin ^{2}(\omega t)>= < \cos ^{2}(\omega t)>=\dfrac{1}{2}\right]\)
PHXI15:WAVES
355185
In the standing wave shown, particles at the positions \(A\) and \(B\) have a phase difference of
1 \(\dfrac{\pi}{2}\)
2 \(0\)
3 \(\pi\)
4 \(\dfrac{5 \pi}{6}\)
Explanation:
Phase difference between any two points present in two adjacent loops is always equal to \(\pi\).
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PHXI15:WAVES
355182
The speed of transverse waves in a stretched string is \({700 {~cm} / {s}}\). It the string is \(2\,m\) long, the frequency with which it resonates in fundamental mode is
1 \({7 / 2 {~Hz}}\)
2 \({7 / 4 {~Hz}}\)
3 14 Hz
4 \({2 / 7 {~Hz}}\)
Explanation:
\({v=700 {~cm} / {s}=7 {~m} / {s}}\) For one loop, \({l=\lambda / 2}\) \({\lambda=2 l=2 \times 2=4 m}\) \({f=\dfrac{v}{\lambda}=7 / 4 H z}\)
PHXI15:WAVES
355183
The expression \(y=a \sin b x \sin \omega t\) represents a stationary wave. The distance between the consecutive nodes is equal to:
1 \(2 \pi / b\)
2 \(\pi / b\)
3 \(1 / b\)
4 \(\pi / 2 b\)
Explanation:
\(y=a \sin b x \sin \omega t\) On comparing with standard equation of stationary wave \(\begin{aligned}& y=2 a \sin \dfrac{2 \pi x}{\lambda} \cdot \sin \omega t, \text { we get } \\& \dfrac{2 \pi x}{\lambda}=b x, \\& \therefore \lambda=\dfrac{2 \pi}{b}\end{aligned}\) The distance between constructive nodes \(=\dfrac{\lambda}{2}=\dfrac{\pi}{b}\)
PHXI15:WAVES
355184
A string of mass \(m\) is fixed at both ends. The Fundamental mode of string is exicted and it has an angular frequency \(\omega\) and the maximum displacement amplitude A.then \(K.E\) for one periodic time is
1 \(\dfrac{1}{2} m A^{2} \omega^{2}\)
2 \(m A^{2} \omega^{2}\)
3 \(\dfrac{1}{4} m A^{2} \omega^{2}\)
4 \(\dfrac{1}{8} m A^{2} \omega^{2}\)
Explanation:
Let \(y=A \sin k x \sin \omega t\) represents the standing wave fixed at both ends. The velocity of a particle is \(v_{p}=\dfrac{\partial y}{\partial t}\) \(v_{p}=A \omega \sin k x \cos \omega t\) The small \(K.E\) of the element is \(\begin{aligned}& d k=\dfrac{1}{2} d m v_{p}^{2}=\dfrac{1}{2} \mu d x v_{p}^{2} \\& d k=\dfrac{1}{2} \mu A^{2} \omega^{2} \sin ^{2} k x \cos ^{2} \omega t d x \\& k=\dfrac{1}{2} \mu A^{2} \omega^{2} \cos ^{2} \omega t \int_{0}^{L} \sin ^{2} k x d x \\& k=\dfrac{1}{4} \mu A^{2} \omega^{2} \cos ^{2} \omega t \int_{0}^{L}(1-\cos 2 k x) d x \\& k=\dfrac{1}{4} \mu A^{2} \omega^{2} \cos ^{2} \omega t\left[L-\dfrac{1}{2 k}[\sin 2 k x]_{0}^{L}\right]\end{aligned}\) given that the string is in Fundamental mode \(\begin{aligned}& \dfrac{\lambda}{2}=L \Rightarrow \lambda=2 L \Rightarrow k=\dfrac{2 \pi}{2 L} \\& k=\dfrac{1}{4} \mu L A^{2} \omega^{2} \cos ^{2} \omega t \\& =\dfrac{1}{4} m A^{2} \omega^{2} \cos ^{2} \omega t \quad(\mu L=m) \\& < k>=\dfrac{1}{4} m A^{2} \omega^{2} < \cos ^{2} \omega t>=\dfrac{1}{8} m A^{2} \omega^{2}\end{aligned}\) (for one time period) \(\left[ < \sin ^{2}(\omega t)>= < \cos ^{2}(\omega t)>=\dfrac{1}{2}\right]\)
PHXI15:WAVES
355185
In the standing wave shown, particles at the positions \(A\) and \(B\) have a phase difference of
1 \(\dfrac{\pi}{2}\)
2 \(0\)
3 \(\pi\)
4 \(\dfrac{5 \pi}{6}\)
Explanation:
Phase difference between any two points present in two adjacent loops is always equal to \(\pi\).
355182
The speed of transverse waves in a stretched string is \({700 {~cm} / {s}}\). It the string is \(2\,m\) long, the frequency with which it resonates in fundamental mode is
1 \({7 / 2 {~Hz}}\)
2 \({7 / 4 {~Hz}}\)
3 14 Hz
4 \({2 / 7 {~Hz}}\)
Explanation:
\({v=700 {~cm} / {s}=7 {~m} / {s}}\) For one loop, \({l=\lambda / 2}\) \({\lambda=2 l=2 \times 2=4 m}\) \({f=\dfrac{v}{\lambda}=7 / 4 H z}\)
PHXI15:WAVES
355183
The expression \(y=a \sin b x \sin \omega t\) represents a stationary wave. The distance between the consecutive nodes is equal to:
1 \(2 \pi / b\)
2 \(\pi / b\)
3 \(1 / b\)
4 \(\pi / 2 b\)
Explanation:
\(y=a \sin b x \sin \omega t\) On comparing with standard equation of stationary wave \(\begin{aligned}& y=2 a \sin \dfrac{2 \pi x}{\lambda} \cdot \sin \omega t, \text { we get } \\& \dfrac{2 \pi x}{\lambda}=b x, \\& \therefore \lambda=\dfrac{2 \pi}{b}\end{aligned}\) The distance between constructive nodes \(=\dfrac{\lambda}{2}=\dfrac{\pi}{b}\)
PHXI15:WAVES
355184
A string of mass \(m\) is fixed at both ends. The Fundamental mode of string is exicted and it has an angular frequency \(\omega\) and the maximum displacement amplitude A.then \(K.E\) for one periodic time is
1 \(\dfrac{1}{2} m A^{2} \omega^{2}\)
2 \(m A^{2} \omega^{2}\)
3 \(\dfrac{1}{4} m A^{2} \omega^{2}\)
4 \(\dfrac{1}{8} m A^{2} \omega^{2}\)
Explanation:
Let \(y=A \sin k x \sin \omega t\) represents the standing wave fixed at both ends. The velocity of a particle is \(v_{p}=\dfrac{\partial y}{\partial t}\) \(v_{p}=A \omega \sin k x \cos \omega t\) The small \(K.E\) of the element is \(\begin{aligned}& d k=\dfrac{1}{2} d m v_{p}^{2}=\dfrac{1}{2} \mu d x v_{p}^{2} \\& d k=\dfrac{1}{2} \mu A^{2} \omega^{2} \sin ^{2} k x \cos ^{2} \omega t d x \\& k=\dfrac{1}{2} \mu A^{2} \omega^{2} \cos ^{2} \omega t \int_{0}^{L} \sin ^{2} k x d x \\& k=\dfrac{1}{4} \mu A^{2} \omega^{2} \cos ^{2} \omega t \int_{0}^{L}(1-\cos 2 k x) d x \\& k=\dfrac{1}{4} \mu A^{2} \omega^{2} \cos ^{2} \omega t\left[L-\dfrac{1}{2 k}[\sin 2 k x]_{0}^{L}\right]\end{aligned}\) given that the string is in Fundamental mode \(\begin{aligned}& \dfrac{\lambda}{2}=L \Rightarrow \lambda=2 L \Rightarrow k=\dfrac{2 \pi}{2 L} \\& k=\dfrac{1}{4} \mu L A^{2} \omega^{2} \cos ^{2} \omega t \\& =\dfrac{1}{4} m A^{2} \omega^{2} \cos ^{2} \omega t \quad(\mu L=m) \\& < k>=\dfrac{1}{4} m A^{2} \omega^{2} < \cos ^{2} \omega t>=\dfrac{1}{8} m A^{2} \omega^{2}\end{aligned}\) (for one time period) \(\left[ < \sin ^{2}(\omega t)>= < \cos ^{2}(\omega t)>=\dfrac{1}{2}\right]\)
PHXI15:WAVES
355185
In the standing wave shown, particles at the positions \(A\) and \(B\) have a phase difference of
1 \(\dfrac{\pi}{2}\)
2 \(0\)
3 \(\pi\)
4 \(\dfrac{5 \pi}{6}\)
Explanation:
Phase difference between any two points present in two adjacent loops is always equal to \(\pi\).
355182
The speed of transverse waves in a stretched string is \({700 {~cm} / {s}}\). It the string is \(2\,m\) long, the frequency with which it resonates in fundamental mode is
1 \({7 / 2 {~Hz}}\)
2 \({7 / 4 {~Hz}}\)
3 14 Hz
4 \({2 / 7 {~Hz}}\)
Explanation:
\({v=700 {~cm} / {s}=7 {~m} / {s}}\) For one loop, \({l=\lambda / 2}\) \({\lambda=2 l=2 \times 2=4 m}\) \({f=\dfrac{v}{\lambda}=7 / 4 H z}\)
PHXI15:WAVES
355183
The expression \(y=a \sin b x \sin \omega t\) represents a stationary wave. The distance between the consecutive nodes is equal to:
1 \(2 \pi / b\)
2 \(\pi / b\)
3 \(1 / b\)
4 \(\pi / 2 b\)
Explanation:
\(y=a \sin b x \sin \omega t\) On comparing with standard equation of stationary wave \(\begin{aligned}& y=2 a \sin \dfrac{2 \pi x}{\lambda} \cdot \sin \omega t, \text { we get } \\& \dfrac{2 \pi x}{\lambda}=b x, \\& \therefore \lambda=\dfrac{2 \pi}{b}\end{aligned}\) The distance between constructive nodes \(=\dfrac{\lambda}{2}=\dfrac{\pi}{b}\)
PHXI15:WAVES
355184
A string of mass \(m\) is fixed at both ends. The Fundamental mode of string is exicted and it has an angular frequency \(\omega\) and the maximum displacement amplitude A.then \(K.E\) for one periodic time is
1 \(\dfrac{1}{2} m A^{2} \omega^{2}\)
2 \(m A^{2} \omega^{2}\)
3 \(\dfrac{1}{4} m A^{2} \omega^{2}\)
4 \(\dfrac{1}{8} m A^{2} \omega^{2}\)
Explanation:
Let \(y=A \sin k x \sin \omega t\) represents the standing wave fixed at both ends. The velocity of a particle is \(v_{p}=\dfrac{\partial y}{\partial t}\) \(v_{p}=A \omega \sin k x \cos \omega t\) The small \(K.E\) of the element is \(\begin{aligned}& d k=\dfrac{1}{2} d m v_{p}^{2}=\dfrac{1}{2} \mu d x v_{p}^{2} \\& d k=\dfrac{1}{2} \mu A^{2} \omega^{2} \sin ^{2} k x \cos ^{2} \omega t d x \\& k=\dfrac{1}{2} \mu A^{2} \omega^{2} \cos ^{2} \omega t \int_{0}^{L} \sin ^{2} k x d x \\& k=\dfrac{1}{4} \mu A^{2} \omega^{2} \cos ^{2} \omega t \int_{0}^{L}(1-\cos 2 k x) d x \\& k=\dfrac{1}{4} \mu A^{2} \omega^{2} \cos ^{2} \omega t\left[L-\dfrac{1}{2 k}[\sin 2 k x]_{0}^{L}\right]\end{aligned}\) given that the string is in Fundamental mode \(\begin{aligned}& \dfrac{\lambda}{2}=L \Rightarrow \lambda=2 L \Rightarrow k=\dfrac{2 \pi}{2 L} \\& k=\dfrac{1}{4} \mu L A^{2} \omega^{2} \cos ^{2} \omega t \\& =\dfrac{1}{4} m A^{2} \omega^{2} \cos ^{2} \omega t \quad(\mu L=m) \\& < k>=\dfrac{1}{4} m A^{2} \omega^{2} < \cos ^{2} \omega t>=\dfrac{1}{8} m A^{2} \omega^{2}\end{aligned}\) (for one time period) \(\left[ < \sin ^{2}(\omega t)>= < \cos ^{2}(\omega t)>=\dfrac{1}{2}\right]\)
PHXI15:WAVES
355185
In the standing wave shown, particles at the positions \(A\) and \(B\) have a phase difference of
1 \(\dfrac{\pi}{2}\)
2 \(0\)
3 \(\pi\)
4 \(\dfrac{5 \pi}{6}\)
Explanation:
Phase difference between any two points present in two adjacent loops is always equal to \(\pi\).
