The pattern of standing waves formed on a stretched string at two instants of time are shown in Fig. The velocity of two waves superimposing to form stationary waves is 360 ms–1 and their frequencies are 256 Hz

(a) Calculate the time at which the second curve is plotted.

(b) Mark nodes and antinodes on the curve.

(c) Calculate the distance between A′ and C′ .

<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>Given frequency of the wave v=256Hz</p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; T=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>256</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>3.9</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mi>s</mi> </math> </span></p><p><strong> (a) </strong>time taken to pass through mean position is</p><p>t=T/4=1/40 =3.9<span contenteditable="false"> <math> <mo>×</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> <mi>s</mi> <mo>=</mo> </math> </span>9.8<span contenteditable="false"> <math> <mo>×</mo> </math> </span>10^-4s</p><p><strong> (b) </strong>nodes are A, B, C, D, E (having zero displacement)</p><p>antinodes are A’ and C’ (having maximum displacement)</p><p><strong> (c) </strong>it is clear from diagram A’ and C’ are forming antinodes have<span contenteditable="false"> <math> <mi>λ</mi> </math> </span> separation= v/v’=360/256=1.41m</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment