201. A wave on a string has speed \(30\,\text{m s}^{-1}\) when produced at \(5.0\,\text{Hz}\). The same string under the same tension is driven at \(10\,\text{Hz}\). The new wavelength is
ⓐ. \(1.5\,\text{m}\)
ⓑ. \(3.0\,\text{m}\)
ⓒ. \(6.0\,\text{m}\)
ⓓ. \(60\,\text{m}\)
Correct Answer: \(3.0\,\text{m}\)
Explanation: \( \textbf{String condition:} \) The same string and same tension mean the wave speed remains \(30\,\text{m s}^{-1}\).
\( \textbf{New frequency:} \)
\[
f'=10\,\text{Hz}=10\,\text{s}^{-1}
\]
\( \textbf{Required quantity:} \) New wavelength \(\lambda'\).
\( \textbf{Wave relation:} \)
\[
v=f'\lambda'
\]
\( \textbf{Rearranging:} \)
\[
\lambda'=\frac{v}{f'}
\]
\( \textbf{Substitution:} \)
\[
\lambda'=\frac{30\,\text{m s}^{-1}}{10\,\text{s}^{-1}}
\]
\( \textbf{Calculation:} \)
\[
\lambda'=3.0\,\text{m}
\]
\( \textbf{Final answer:} \) The new wavelength is \(3.0\,\text{m}\).
The frequency has changed, but the speed remains fixed because the string conditions are unchanged.
202. The derivation of \(v=\sqrt{\frac{T_s}{\mu}}\) for a stretched string mainly depends on the idea that
ⓐ. tension supplies the restoring force for a small string element
ⓑ. the string particles move permanently along the string with the wave
ⓒ. the wave speed is fixed only by the frequency of the oscillator
ⓓ. gravity must be the only force acting on every element of the string
Correct Answer: tension supplies the restoring force for a small string element
Explanation: In the usual derivation for a wave on a stretched string, a small element of the string is considered. When the string is slightly curved by a transverse disturbance, the tensions at the two ends of the small element have components that provide a restoring force. This restoring force produces transverse acceleration of the string element. The inertia involved is measured by the mass per unit length, \(\mu\). Combining the restoring effect of tension with the inertia of the string gives \(v=\sqrt{\frac{T_s}{\mu}}\). The derivation is based on small transverse disturbances, not on permanent transport of string material.
203. A dimensional check is applied to \(v=\sqrt{\frac{T_s}{\mu}}\). Since \(T_s\) has unit \(\text{N}\) and \(\mu\) has unit \(\text{kg m}^{-1}\), the unit of \(\frac{T_s}{\mu}\) is
ⓐ. \(\text{m}^2\text{s}^{-1}\)
ⓑ. \(\text{kg m}^{-1}\text{s}^{-2}\)
ⓒ. \(\text{s}^{-2}\)
ⓓ. \(\text{m}^2\text{s}^{-2}\)
Correct Answer: \(\text{m}^2\text{s}^{-2}\)
Explanation: \( \textbf{Tension unit:} \)
\[
\text{N}=\text{kg m s}^{-2}
\]
\( \textbf{Linear mass density unit:} \)
\[
\mu=\text{kg m}^{-1}
\]
\( \textbf{Ratio unit:} \)
\[
\frac{T_s}{\mu}=\frac{\text{kg m s}^{-2}}{\text{kg m}^{-1}}
\]
\( \textbf{Canceling \(\text{kg}\):} \)
\[
\frac{T_s}{\mu}=\text{m s}^{-2}\times\text{m}
\]
\( \textbf{Simplification:} \)
\[
\frac{T_s}{\mu}=\text{m}^2\text{s}^{-2}
\]
\( \textbf{Taking square root:} \)
\[
\sqrt{\text{m}^2\text{s}^{-2}}=\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The unit of \(\frac{T_s}{\mu}\) is \(\text{m}^2\text{s}^{-2}\).
The square root then correctly gives the unit of wave speed.
204. The formula \(v=\sqrt{\frac{T_s}{\mu}}\) is used for a stretched string most safely when the wave is
ⓐ. a small transverse disturbance on a uniform string
ⓑ. a sound wave travelling through vacuum
ⓒ. a large permanent flow of string material
ⓓ. a wave whose speed is controlled only by amplitude
Correct Answer: a small transverse disturbance on a uniform string
Explanation: The standard string-wave formula assumes a stretched string with uniform linear mass density. The disturbance is treated as small so that the string tension remains nearly constant and the usual small-slope approximation is valid. The wave is transverse because the string elements move perpendicular to the direction of propagation. The formula is not for sound in vacuum, because sound needs a medium and the string formula uses string tension and linear density. It is also not a formula for bulk motion of string material. The physical conditions behind the relation matter as much as the algebraic form.
205. A string of length \(2.0\,\text{m}\) has mass \(0.040\,\text{kg}\). It is stretched with tension \(80\,\text{N}\). The transverse wave speed on the string is
ⓐ. \(20\,\text{m s}^{-1}\)
ⓑ. \(40\,\text{m s}^{-1}\)
ⓒ. \(63\,\text{m s}^{-1}\)
ⓓ. \(4000\,\text{m s}^{-1}\)
Correct Answer: \(63\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given data:} \) Length \(L=2.0\,\text{m}\), mass \(m=0.040\,\text{kg}\), tension \(T_s=80\,\text{N}\).
\( \textbf{First find linear mass density:} \)
\[
\mu=\frac{m}{L}
\]
\( \textbf{Substitution for \(\mu\):} \)
\[
\mu=\frac{0.040\,\text{kg}}{2.0\,\text{m}}=0.020\,\text{kg m}^{-1}
\]
\( \textbf{Wave speed relation:} \)
\[
v=\sqrt{\frac{T_s}{\mu}}
\]
\( \textbf{Substitution:} \)
\[
v=\sqrt{\frac{80}{0.020}}
\]
\( \textbf{Intermediate value:} \)
\[
\frac{80}{0.020}=4000
\]
\( \textbf{Square root:} \)
\[
v=\sqrt{4000}\approx63\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The wave speed is approximately \(63\,\text{m s}^{-1}\).
The mass must first be converted into mass per unit length before using the string-speed formula.
206. A musician wants to double the wave speed on a string without changing the string itself. The required change in tension is
ⓐ. make the tension twice
ⓑ. make the tension four times
ⓒ. make the tension half
ⓓ. make the tension one-fourth
Correct Answer: make the tension four times
Explanation: \( \textbf{String unchanged:} \) The linear mass density \(\mu\) remains constant.
\( \textbf{Speed relation:} \)
\[
v=\sqrt{\frac{T_s}{\mu}}
\]
\( \textbf{Required condition:} \)
\[
v'=2v
\]
\( \textbf{Using proportionality:} \)
\[
v\propto\sqrt{T_s}
\]
\( \textbf{Ratio form:} \)
\[
\frac{v'}{v}=\sqrt{\frac{T_s'}{T_s}}
\]
\( \textbf{Substitution:} \)
\[
2=\sqrt{\frac{T_s'}{T_s}}
\]
\( \textbf{Squaring:} \)
\[
4=\frac{T_s'}{T_s}
\]
\( \textbf{Final answer:} \) The tension must be made four times.
Doubling speed requires quadrupling tension because the relation contains a square root.
207. The following statements refer to transverse waves on stretched strings.
I. Increasing \(T_s\) increases wave speed if \(\mu\) is constant.
II. Increasing \(\mu\) increases wave speed if \(T_s\) is constant.
III. In the same string under the same tension, changing frequency changes wavelength rather than wave speed.
The supported statements are
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I, II, and III
ⓓ. I and III only
Correct Answer: I and III only
Explanation: Statement I is supported by \(v=\sqrt{\frac{T_s}{\mu}}\), because increasing \(T_s\) increases the speed. Statement II is not supported because increasing \(\mu\) increases inertia per unit length and decreases speed. Statement III is supported because, in the same string under the same tension, \(T_s\) and \(\mu\) remain unchanged. The speed therefore remains fixed while the relation \(v=f\lambda\) makes wavelength adjust when frequency changes. This is a common place where source-controlled frequency and medium-controlled speed must be kept separate. The correct set keeps both the formula and the physical meaning consistent.
208. Two strings have the same length and are under the same tension. String R is made of a denser material, so its mass per unit length is larger. A pulse sent along string R will travel
ⓐ. faster than on the lighter string
ⓑ. slower than on the lighter string
ⓒ. with the same speed regardless of mass per unit length
ⓓ. only if the source frequency is zero
Correct Answer: slower than on the lighter string
Explanation: The speed of a transverse pulse on a stretched string is \(v=\sqrt{\frac{T_s}{\mu}}\). The two strings have the same tension, so the difference comes from \(\mu\). A denser or heavier string has larger mass per unit length. Larger \(\mu\) means more inertia must be accelerated by the restoring force due to tension. Therefore, the wave speed is smaller on the heavier string. The pulse can still travel, but it travels more slowly because the string has more inertia per unit length.
209. A graph is plotted between \(v^2\) and \(T_s\) for transverse waves on a string of fixed linear mass density \(\mu\). The graph is a straight line through the origin. Its slope is
ⓐ. \(\mu\)
ⓑ. \(\sqrt{\mu}\)
ⓒ. \(\frac{1}{\sqrt{\mu}}\)
ⓓ. \(\frac{1}{\mu}\)
Correct Answer: \(\frac{1}{\mu}\)
Explanation: \( \textbf{Starting relation:} \)
\[
v=\sqrt{\frac{T_s}{\mu}}
\]
\( \textbf{Squaring both sides:} \)
\[
v^2=\frac{T_s}{\mu}
\]
\( \textbf{Writing in graph form:} \)
\[
v^2=\left(\frac{1}{\mu}\right)T_s
\]
\( \textbf{Comparison with straight-line form:} \) If \(v^2\) is plotted on the vertical axis and \(T_s\) on the horizontal axis, the slope is \(\frac{1}{\mu}\).
\( \textbf{Intercept:} \) The ideal relation passes through the origin.
\( \textbf{Final answer:} \) The slope is \(\frac{1}{\mu}\).
This graph form removes the square root and makes the inverse dependence on linear mass density visible.
210. A string has \(\mu=0.025\,\text{kg m}^{-1}\). The measured slope of a \(v^2\) versus \(T_s\) graph should be
ⓐ. \(0.025\,\text{kg m}^{-1}\)
ⓑ. \(0.158\,\text{kg}^{1/2}\text{m}^{-1/2}\)
ⓒ. \(40\,\text{kg}^{-1}\text{m}\)
ⓓ. \(25\,\text{m s}^{-1}\)
Correct Answer: \(40\,\text{kg}^{-1}\text{m}\)
Explanation: \( \textbf{Given data:} \) Linear mass density \(\mu=0.025\,\text{kg m}^{-1}\).
\( \textbf{Graph relation:} \)
\[
v^2=\left(\frac{1}{\mu}\right)T_s
\]
\( \textbf{Slope of \(v^2\) versus \(T_s\):} \)
\[
\text{slope}=\frac{1}{\mu}
\]
\( \textbf{Substitution:} \)
\[
\text{slope}=\frac{1}{0.025\,\text{kg m}^{-1}}
\]
\( \textbf{Calculation:} \)
\[
\text{slope}=40\,\text{kg}^{-1}\text{m}
\]
\( \textbf{Unit check:} \) Since \(\frac{v^2}{T_s}=\frac{\text{m}^2\text{s}^{-2}}{\text{kg m s}^{-2}}=\text{kg}^{-1}\text{m}\), the unit is consistent.
\( \textbf{Final answer:} \) The slope is \(40\,\text{kg}^{-1}\text{m}\).
The slope is the reciprocal of \(\mu\), not \(\mu\) itself.
211. Linear mass density \(\mu\) of a string is defined as
ⓐ. \(\mu=\frac{L}{m}\), with unit \(\text{m kg}^{-1}\)
ⓑ. \(\mu=\frac{m}{L}\), with unit \(\text{kg m}^{-1}\)
ⓒ. \(\mu=mL\), with unit \(\text{kg m}\)
ⓓ. \(\mu=\frac{T_s}{v}\), with unit \(\text{N s m}^{-1}\)
Correct Answer: \(\mu=\frac{m}{L}\), with unit \(\text{kg m}^{-1}\)
Explanation: Linear mass density tells how much mass is present per unit length of a string. If a string has mass \(m\) and length \(L\), then \(\mu=\frac{m}{L}\). Its SI unit is \(\text{kg m}^{-1}\). This quantity is important because a string with larger \(\mu\) has greater inertia per unit length. In the wave-speed relation \(v=\sqrt{\frac{T_s}{\mu}}\), larger \(\mu\) gives smaller speed for the same tension. The reciprocal \(\frac{L}{m}\) would describe length per unit mass, not mass per unit length.
212. A string of length \(5.0\,\text{m}\) has total mass \(0.10\,\text{kg}\). Its linear mass density is
ⓐ. \(0.020\,\text{kg m}^{-1}\)
ⓑ. \(0.50\,\text{kg m}^{-1}\)
ⓒ. \(5.0\,\text{kg m}^{-1}\)
ⓓ. \(50\,\text{kg m}^{-1}\)
Correct Answer: \(0.020\,\text{kg m}^{-1}\)
Explanation: \( \textbf{Given data:} \) Mass \(m=0.10\,\text{kg}\), length \(L=5.0\,\text{m}\).
\( \textbf{Required quantity:} \) Linear mass density \(\mu\).
\( \textbf{Definition:} \)
\[
\mu=\frac{m}{L}
\]
\( \textbf{Substitution:} \)
\[
\mu=\frac{0.10\,\text{kg}}{5.0\,\text{m}}
\]
\( \textbf{Calculation:} \)
\[
\mu=0.020\,\text{kg m}^{-1}
\]
\( \textbf{Final answer:} \) The linear mass density is \(0.020\,\text{kg m}^{-1}\).
The unit must be mass per length, so \(\text{kg m}^{-1}\) confirms the correct ratio.
213. A string is made thicker using the same material and is kept under the same tension as before. For transverse waves on the string, the speed is expected to
ⓐ. increase because the string is thicker
ⓑ. decrease because linear mass density is larger
ⓒ. remain unchanged because the material is the same
ⓓ. become independent of tension
Correct Answer: decrease because linear mass density is larger
Explanation: The speed of a transverse wave on a stretched string is \(v=\sqrt{\frac{T_s}{\mu}}\). If the same material is made into a thicker string, the mass per unit length \(\mu\) increases. With the same tension \(T_s\), a larger \(\mu\) gives a smaller value of \(\frac{T_s}{\mu}\). Therefore, the wave speed decreases. The material being the same does not mean the linear mass density is unchanged, because thickness affects mass per unit length. The physical reason is that a heavier string has more inertia for the same restoring tension.
214. A transverse wave travels at \(50\,\text{m s}^{-1}\) on a string of linear mass density \(0.020\,\text{kg m}^{-1}\). The tension in the string is
ⓐ. \(25\,\text{N}\)
ⓑ. \(100\,\text{N}\)
ⓒ. \(250\,\text{N}\)
ⓓ. \(50\,\text{N}\)
Correct Answer: \(50\,\text{N}\)
Explanation: \( \textbf{Given data:} \) Wave speed \(v=50\,\text{m s}^{-1}\), linear mass density \(\mu=0.020\,\text{kg m}^{-1}\).
\( \textbf{Required quantity:} \) Tension \(T_s\).
\( \textbf{String-wave relation:} \)
\[
v=\sqrt{\frac{T_s}{\mu}}
\]
\( \textbf{Squaring both sides:} \)
\[
v^2=\frac{T_s}{\mu}
\]
\( \textbf{Rearranging:} \)
\[
T_s=\mu v^2
\]
\( \textbf{Substitution:} \)
\[
T_s=(0.020)(50)^2
\]
\( \textbf{Calculation:} \)
\[
(50)^2=2500
\]
\[
T_s=0.020\times2500=50\,\text{N}
\]
\( \textbf{Final answer:} \) The string tension is \(50\,\text{N}\).
The speed must be squared before multiplying by \(\mu\).
215. Two equal-length strings are stretched with the same tension. String P has mass \(0.050\,\text{kg}\), and string Q has mass \(0.200\,\text{kg}\). The speed of waves on Q compared with P is
ⓐ. four times as large
ⓑ. two times as large
ⓒ. one-half as large
ⓓ. one-fourth as large
Correct Answer: one-half as large
Explanation: \( \textbf{Same length condition:} \) For equal lengths, linear mass density is proportional to total mass.
\( \textbf{Mass ratio:} \)
\[
\frac{\mu_Q}{\mu_P}=\frac{0.200}{0.050}=4
\]
\( \textbf{Same tension relation:} \)
\[
v=\sqrt{\frac{T_s}{\mu}}
\]
\( \textbf{Speed ratio:} \)
\[
\frac{v_Q}{v_P}=\sqrt{\frac{\mu_P}{\mu_Q}}
\]
\( \textbf{Substitution:} \)
\[
\frac{v_Q}{v_P}=\sqrt{\frac{1}{4}}
\]
\( \textbf{Calculation:} \)
\[
\frac{v_Q}{v_P}=\frac{1}{2}
\]
\( \textbf{Final answer:} \) The speed on string Q is one-half the speed on string P.
The fourfold increase in mass per unit length gives a twofold decrease in speed because of the square root.
216. A record for a stretched string says: \(m=0.060\,\text{kg}\), \(L=3.0\,\text{m}\), and \(T_s=45\,\text{N}\). The correct sequence for finding the wave speed is
ⓐ. use \(v=\sqrt{\frac{T_s}{m}}\) before finding \(\mu\)
ⓑ. find \(\mu\), then use \(v=\sqrt{\frac{T_s}{\mu}}\)
ⓒ. first find \(f=\frac{1}{T_s}\), then use \(v=fL\)
ⓓ. use \(v=\frac{T_s}{\mu}\) after finding \(\mu\)
Correct Answer: find \(\mu\), then use \(v=\sqrt{\frac{T_s}{\mu}}\)
Explanation: The string-speed formula uses linear mass density \(\mu\), not total mass \(m\) directly. Therefore, the first step is to calculate \(\mu=\frac{m}{L}\). After that, the wave speed is found from \(v=\sqrt{\frac{T_s}{\mu}}\). Using \(m\) in place of \(\mu\) would ignore the length of the string and give a wrong unit. Removing the square root would also make the result dimensionally wrong for speed. The formula connects restoring tension with inertia per unit length, so both \(T_s\) and \(\mu\) must be used in their correct roles.
217. A sound wave cannot travel through a perfect vacuum because
ⓐ. no particles pass on pressure disturbances
ⓑ. its wavelength becomes larger than any possible distance
ⓒ. its frequency must become exactly zero
ⓓ. electromagnetic waves block sound waves in vacuum
Correct Answer: no particles pass on pressure disturbances
Explanation: Sound is a mechanical wave, so it needs a material medium for propagation. In air, sound travels through compressions and rarefactions of air particles. A perfect vacuum has no particles that can be compressed or rarefied. Without particles, there is no mechanism to pass the disturbance from one region to another. The frequency of the source may still exist as vibration of the source, but the sound wave cannot propagate through the vacuum. This separates sound from electromagnetic waves such as light, which can travel through vacuum.
218. The speed of sound in a material medium depends mainly on
ⓐ. the colour of the medium and the shape of the source
ⓑ. the distance from the listener only
ⓒ. the loudness of the sound only
ⓓ. the elasticity of the medium and its inertia
Correct Answer: the elasticity of the medium and its inertia
Explanation: Sound travels because the medium can be compressed and then restore itself. The elastic property of the medium provides the restoring effect. The density or inertia of the medium resists rapid motion of its particles. Sound speed depends on the balance between these two features. A medium with strong restoring ability and relatively low inertia tends to transmit sound faster. Loudness and distance affect the received intensity, but they do not by themselves define the basic speed of sound in the medium.
219. Sound generally travels faster in solids than in gases because solids usually have
ⓐ. particles with no restoring forces
ⓑ. stronger elastic restoring forces
ⓒ. zero density
ⓓ. no need for particle interaction
Correct Answer: stronger elastic restoring forces
Explanation: Sound propagation requires neighbouring particles of the medium to interact and pass on a disturbance. Solids generally have particles that are strongly connected, so a disturbance can be transmitted rapidly. Although solids may have high density, their elastic restoring forces are usually much stronger than those in gases. This often makes sound speed greater in solids than in liquids and gases. A gas has weaker elastic restoring response under ordinary conditions. The speed comparison depends on both elasticity and inertia, not on density alone.
220. A bell rings inside a closed jar. As air is gradually removed, the sound heard outside becomes weaker, although the bell is still seen vibrating. This observation shows that
ⓐ. sound needs a material medium for propagation
ⓑ. light needs air more than sound does
ⓒ. the bell stops vibrating immediately in vacuum
ⓓ. sound is an electromagnetic wave
Correct Answer: sound needs a material medium for propagation
Explanation: The vibrating bell acts as the source of the sound. When air is present, air particles carry the pressure disturbance from the bell to the outside. As air is removed, fewer particles are available to transmit the disturbance, so the sound becomes weaker. The bell may still be visible because light can travel through the glass and evacuated region. This contrast shows that sound is mechanical, while light is electromagnetic. The decreasing sound is therefore due to removal of the medium, not due to disappearance of the source vibration.
